
Class _.' ; •._: 

Rnift - 

Gopightfl 

COPYRIGHT DEBOam 



2-3 



TRIGONOMETRY, SURVEYING 
AND NAVIGATION 



BY 



G. A. WENTWORTH 

Author of a Series of Text-Books is Mathematics 



SECOND REVISED EDITION 



BOSTON, U.S.A. 

GINN & COMPANY, PUBLISHERS 

(Re Sttbenaettm JJrega 

1903 



THE LIBRARY OF 
CONGRESS, 

Two Copies Receive 

SEP 26 1W> 

Copyright fcntf) 

CLASS CL/ XXc N 

COPY B. 






GC 






MATHEMATICAL TEXT-BOOKS 

BY 

GEORGE A. WENT WORTH 



Elementary Arithmetic 

Practical Arithmetic 

Mental Arithmetic 

Primary Arithmetic (Wentworth and Reed) 

Grammar School Arithmetic 

Advanced Arithmetic 

Exercises in Arithmetic (Wentworth and Hill) 

First Steps in Algebra 

School Algebra 

New School Algebra 

Higher Algebra 

Elements of Algebra 

Complete Algebra 

Shorter Course in Algebra 

College Algebra (Revised Edition) 

Exercises in Algebra (Wentworth and Hill) 

First Steps in Geometry (Wentworth and Hill) 

Plane and Solid Geometry (Revised) 

Plane Geometry (Revised) 

Solid Geometry (Revised) 

Plane and Solid Geometry and Plane Trigonometry 

(Second Revised Edition) 
Analytic Geometry 
Logarithms, Metric Measures, etc. 
Geometrical Exercises 
Syllabus of Geometry 

Examination Manual in Geometry (Wentworth and Hill) 
Exercise Manual in Geometry (wentworth and Hill) 
Plane Trigonometry (Second Revised Edition) 
Plane Trigonometry and Tables (Second Revised 

Edition) 
Plane and Spherical Trigonometry (Second Revised 

Edition) 
Plane and Spherical Trigonometry and Tables 

(Second Revised Edition) 
Plane Trigonometry, Surveying, and Tables (Second 

Revised Edition) 
Surveying and Tables (Second Revised Edition) 
Plane and Spherical Trigonometry, Surveying, and 

Tables. (Second Rav'.sed Edition) 
Plfcnfc ;and; Spherjc'di; .Trigonometry, Surveying, and 

Navigation (Second Revised Edition) 
Logarithmic and f Trigonometric Tables 
1 Seven Tables (Wentworth and Hill) 

; . ,' <C|)ttj)lcte »' ■ 



Copyright, 1882, 1895, 1896, 1902, 1903, by 
G. A. WENTWORTH 



ALL RIGHTS RESERVED 



PREFACE 

In preparing this work the aim has been to furnish just so much 
of Trigonometry as is actually taught in our best schools and col- 
leges. Consequently, all investigations that are important only for 
the special student have been omitted, except the development of 
functions in series. The principles have been unfolded with the 
utmost brevity consistent with simplicity and clearness, and inter- 
esting problems have been selected with a view to awaken a real love 
for the study. Much time and labor have been spent in devising 
the simplest proofs for the propositions, and in exhibiting the best 
methods of arranging the logarithmic work. 

The object of the work on Surveying and on Navigation is to 
present these subjects clearly and in so small a compass that students 
in general may find the time to acquire a competent knowledge of 
these important studies. 

The author is under obligation to G. A. Hill, A.M., of Cam- 
bridge, Mass.; to Professor J. L. Patterson, of Chestnut Hill. Pa.: 
to Dr. F. N. Cole, of New York, N.Y.; to Professor S. F. Norris, of 
Baltimore, Md. ; and to Professor B. F. Yanney, of Alliance. Ohio. 
Professor Yanney has done most of the work in the revision of the 
Surveying. Miss M. Gertrude Cross, of Boston, Mass., has furnished 
the drawings for this edition. 

G. A. WEXTWORTH. 

Exeter, N.H., 1903. 



CONTENTS 

[The numbers refer to the pages.] 

PLANE TRIGONOMETRY 

CHAPTER I. Trigonometric Functions of Acute Angles : 

Angular measure, 1; trigonometric functions, 3; representation 
of the functions by lines, 7 ; changes in the functions as the angle 
changes, 10; functions of complementary angles, 11; relations of 
the functions of an angle, 13 ; formulas for finding all the other 
functions of an angle when one function of the angle is given, 15 ; 
functions of 45°, 17° ; functions of 30° and 60°, 18. 

CHAPTER II. The Right Triangle : 

Given parts of a right triangle, 20. Solutions without logarithms, 
20 ; Case I, when an acute angle and the hypotenuse are given, 20 
Case II, when an acute angle and the opposite leg are given, 21 
Case III, when an acute angle and an adjacent leg are given, 21 
Case IV, when the hypotenuse and a leg are given, 22 ; Case V, 
when the two legs are given, 22 ; general method of solving the right 
triangle, 23 ; solutions by logarithms, 25 ; area of the right triangle, 
27 ; the isosceles triangle, 32; the regular polygon, 34. 

CHAPTER III. Goniometry: 

Definition of goniometry, 36 ; positive and negative quantities, 36 ; 
co-ordinates of a point in a plane, 37 ; angles of any magnitude, 38 ; 
functions of any angle, 40; functions of a variable angle, 42 ; func- 
tions of angles larger than 360°, 44 ; extension of formulas for acute 
angles to angles of any magnitude, 44 ; reduction of the functions of 
all angles to the functions of angles in the first quadrant, 47 ; func- 
tions of angles that differ by 90°, 50 ; functions of a negative angle, 
51 ; functions of the sum of two angles, 53 ; functions of the differ- 
ence of two angles, 56 ; functions of twice an angle, 58 ; functions 
of half an angle, 58 ; sums and differences of functions, 59 ; anti- 
trigonometric functions, 61. 

v 



vi CONTENTS 

CHAPTER IV. The Oblique Triangle : 

Law of sines, 64 ; law of cosines, 66 ; law of tangents, 67. Solu- 
tions : Case I, when one side and two angles are given, 69 ; Case II, 
when two sides and the angle opposite one of them are given, 71 ; 
Case III, when two sides and the included angle are given, 76; 
Case IV, when the three sides are given, 80 ; area of a triangle, 85. 

CHAPTER V. Miscellaneous Examples : 

Problems in Plane Trigonometry, 89 ; right triangles, 90 ; oblique 
triangles, 93 ; areas, 98 ; plane sailing, 101 ; parallel and middle lati- 
tude sailing, 103; traverse sailing, 106; examples in goniometry, 
107 ; solution of single equations, 112 ; systems of equations, 116. 

CHAPTER VI. Construction op Tables: 

Logarithms, 119 ; exponential and logarithmic series, 122 ; trigo- 
nometric functions of small angles, 127 ; Simpson's method of con- 
structing a trigonometric table, 129 ; De Moivre's theorem, 131 ; 
expansion of sin x, cos x, and tan x, in infinite series, 135. 



SPHERICAL TRIGONOMETRY 



CHAPTER VII. The Right Spherical Triangle : 

Introduction, 139 ; formulas relating to right spherical triangles, 
143 ; Napier's rules, 146. Solutions : Case I, when the two legs are 
given, 148 ; Case II, when the hypotenuse and a leg are given, 149 ; 
Case III, when a leg and the opposite angle are given, 149 ; Case IV, 
when a leg and an adjacent angle are given, 150 ; Case V, when the 
hypotenuse and an oblfoue angle are given, 150 ; Case VI, when the 
two oblique angles aregiven, 151. The isosceles spherical triangle, 156. 

CHAPTER VIII. The Oblique Spherical Triangle : 

Fundamental formulas, 158; formulas for the half angles and 
sides, 160 ; Gauss's Equations and Napier's Analogies, 163. Solutions : 
Case I, when two sides and the included angle are given, 165 ; Case II, 
when two angles and the included side are given, 168 ; Case III, when 
two sides and an angle opposite one of them are given, 170 ; Case IV, 
when two angles and a side opposite one of them are given, 172 ; 



CONTEXTS vii 

Case V, when the three sides are given, 173 ; Case VI, when the 
three angles are given, 175. Area of a spherical triangle, 177. 

CHAPTER IX. Applications of Spherical Trigonometry : 

To reduce to the horizon an angle measured in space, 181 ; to find 
the distance between two places on the earth's surface when the 
latitudes of the places and the difference of their longitudes are 
given, 182 ; the celestial sphere, 183 ; spherical co-ordinates, 186 ; 
the astronomical triangle, 189 ; to find the declination and the hou- 
angle of a star when the latitude of the observer and the altituc J s 
and the azimuth of the star are given, 191 ; to find the hour angle 
of a star when its declination, its altitude, and the latitude of the 
place are given, 192 ; to find the altitude and the azimuth of a star 
when its declination, its hour angle, and the latitude of the place are 
given, 193 ; to find the latitude of the place when the altitude of a 
star, its declination, and its hour angle are given, 194 ; to find the 
latitude and the longitude of a star when its declination, its right 
ascension, and the obliquity of the ecliptic are given, 195. 

FORMULAS. 

Plane Trigonometry, 202 ; Spherical Trigonometry, 204. 



SURVEYING 
CHAPTER I. Field Instruments: 

Definitions, 209; classification, 209; operations comprised, 210; 
the surveyor's chain, 211 ; the engineer's chain, 211 ; accompanying 
pieces, 212 ; how to chain, 212 ; special constructions by means of the 
chain, 213 ; obstacles to chaining, 215 ; + he tape, 217 ; the compass, 
218 ; kinds of compasses, 219 ; bearing o* o, line, 220 ; checking bear- 
ings, 221 ; obstacles, 222 ; measurement of horizontal angles, 222 ; 
measurement of vertical angles, 223 ; verniers, 223 ; uses of the com- 
pass vernier, 225 ; magnetic declination, 227 ; surveyor's transit, 231 ; 
uses, 232 ; measurement of horizontal angles, 234 ; measurement of 
vertical angles, 234 ; stadia measurements, 234 ; the solar compass, 
236 ; to establish a true meridian, 240 ; the Y level, 244 ; the level- 
ing rod, 244-; substitutes for the Y level, 247 ; the plane table, 248 ; 
to orient the table, 250 ; to plot any point, 251 ; to plot a field, 251 ; 
the three-point problem, 252. 



viii CONTENTS 

CHAPTER II. Office Instruments : 

Definitions, 254 ; the diagonal scale, 254 ; the circular protractor, 
255 ; constructions, 256 ; the planimeter, 257 ; the slide rule, 257. 

CHAPTER III. Land Surveying: 

Definitions, 258; special methods of surveying, and of computing 
areas, 259; rectangular system of co-ordinates, 260; general method 
for farm surveys, 265 ; field notes, 266 ; computation of the area, 
266 ; balancing the work, 268 ; supplying omissions, 269 ; to make 
a plot, 271 ; modification of the latitude and departure method, 
274 ; location surveys, 275 ; illustrative problems, 275 ; laying out 
the public lands, 279 ; reference lines, 279 ; townships, 279 ; sub- 
division of townships, 281 ; meander lines, 281. 

CHAPTER IV. Triangulation: 

Definitions, 282 ; classification, 283 ; measurement of base lines, 
283 ; measurement of angles, 284. 

CHAPTER V. Leveling: 

Definitions, 285; corrections for curvature and refraction, 285; 
differential leveling, 286; single setting of the level, 286; several 
settings of the level, 287 ; profile leveling, 288 ; field work, 289 ; 
making the profile, 292; topographic leveling, 293; drainage sur- 
veying, 294 ; field work, 294 ; plot and profile, 294. 

CHAPTER VI. Railroad Surveying : 

Laying out the route, 297 ; establishing the roadbed, 297 ; exca- 
vations, 297 ; embankments, 298 ; curves, 299 ; methods of laying 
out the curve, 300. 

CHAPTER VII. City Surveying: 

Field-work instruments, 302 ; streets, 302 ; blocks and lots, 304 ; 
plots, 304 ; records, 304. 



THE RIGHT TRIANGLE 


a = 40, b = 27. 


a 2 = 1600 


|o = 1.4815 
tan ,4 =1.4815 
' .'.A = 55° 59' 


£ 2 = 729 

c 2 = 2329 

.-.-c = V2329 


£ = 34°1' 


= 48.26 


SECTION XII 



23 



GENERAL METHOD OF SOLVING THE RIGHT TRIANGLE 



From these five cases it appears that the general method of 
finding an unknown part in a right triangle is as follows : 

Choose from the equation A + B = 90°, and the equations 
that define the functions of the angles, an equation in which 
the required part only is unknown; solve this equation, if 
necessary, to find the value of the unknown part ; then com- 
pute the value. 

Note. In Case IV, if the given sides (here a and c) are nearly alike 
in value, then A is near 90°, and its value cannot be accurately found 
from the tables, because the sines of large angles differ little in value 
(as is evident from Fig. 4). In this case it is better to find B first, by 
means of the formula given on page 59, namely, 



tan f B = 

Example. Given a = 49, c = 50 
1, c + a = 99. 

= 0.01010 



jc — a 

\c + a 



+ 
find A, B, b. 



c — a 
c — a 



c + a 

J C -^ = 0.1005 
\c + a 

tani£ = 0.1005 

.-. i B = 5° 44' 

B = 11° 28' 
A = 78° 32' 



c — a = 1 
c + a = 99 
2 _ a 2 = 99 
& 2 = 99 
6 = V99 
= 9.95 



24 



PLANE TRIGONOMETRY 



EXERCISE VIII 

1. In Case II give another way of finding c, after b has 
been found. 

2. In Case III give another way of finding c, after a has 
been found. 

3. In Case IV give another way of finding b, after the 
angles have been found. 

4. In Case V give another way of finding c, after the 
angles have been found. 

5. Given B and c ; find A, a, b. 

6. Given B and b ; find A, a, c. 

7. Given B and a ; find A, b, c. 

8. Given b and c ; find A, B, a. 

Solve the following right triangles : 



9 


Given 


Required 


a = 3, 


6 = 4. 


^L = 36° 52', 


jB = 53° 8', 


c = 5. 


10 


a— 7, 


c = 13. 


A = 32° 35', 


B = 57° 25', 


6 = 10.954. 


11 


a = 5.3, 


A = 12° 17'. 


B = 77° 43', 


6 = 24.342, 


c = 24.918. 


12 


a = 10.4, 


B = 43° 18'. 


A = 46° 42', 


6 = 9.800, 


c = 14.290. 


13 


c =26, 


^L = 37° 42'. 


£ = 52° 18', 


a = 15.900, 


6 = 20.572. 


14 


c = 140, 


B = 24° 12'. 


A = 65° 48', 


a = 127.694, 


6 = 57.386. 


15 


6 = 19, 


c = 23. 


A = 34° 18', 


B = 55° 42', 


a = 12.961. 


16 


6 = 98, 


c = 135.2. 


A = 43° 33', 


B = 46° 27', 


a = 93.139. 


17 


6 = 42.4, 


A = 32° 14'. 


B = 57° 46', 


a = 26.733, 


c = 50.124. 


18 


6 = 200, 


B = 46° if'. 


^L = 43° 49', 


a = 191.900, 


c = 277.160. 


19 


a = 95, 


6 = 37. 


A = 68° 43', 


B =21° 17', 


c = 101.951. 


20 


a = 6, 


c = 103. 


^1=3° 21', 


E = 86° 39', 


6 = 102.825. 


21 


a = 3.12, 


J5 = 5° 8'. 


A = 84° 52', 


6 = 0.280, 


c = 3.133. 


22 


a = 17, 


c = 18. 


A = 70° 48', 


B = 19° 12', 


6 = 5.916. 


23 


c= 57, 


A = 38° 29'. 


B = 51° 31', 


a = 35.471, 


6 = 44.620. 


24 


a + c = 18, 


6 = 12. 


A = 22° 37', 


5 = 67° 23', 


a = 5, c=13. 


25 


a + 6 = 9, 


c = 8. 


A = 82° 18', 


JB=7°42', - 


'a = 7.928, 
6 = 1.072. 



THE RIGHT TRIANGLE 



25 



SECTION XIII 



SOLUTION BY LOGARITHMS 

Case I 
Given A = 34° 28', c = 18.75 ; find B, a, ft. 



1. 


B = 90° - .4 = 55° 32'. 






2. 


- = sm vl ; .'. a = c sin A. 

c 






3 


- = cos A ; .'. b = c cos ^4. 
c 




^3i°28' 






b 








Fig. 15 



log a = log c + log sin ^ I 
log c == 1.27300 
log sin A = 9.75276 - 10 
log a = 1.02576 
a = 10.611 



log b = log c + log cos A 
log c = 1.27300 
log cos .4 = 9.91617-10 
log b = 1.18917 
b = 15.459 



Case II 
Given A = 62° 10', a = 78 ; find B, b, c. 



1. B = 90°-A= 27° 50'. 






so 


2. - = cot A ; .'. b = a cot .4. 


V 


8 


3. -=sin.4. 




/\62°10' 




c 


1 


■1 ' * 


O 


• >• j a 


^ b 


. . a — c sin vi, ana. c — . 

sin j 


Fig. 16 


log ft = log a -f- log cot .4 


logc 


= log a + colog sin J 


log a = 1.89209 


log a 


= 1.89209 


I cot 4 = 9.72262 - 10 


colog sin A 


= 0.05340 


log ft = 1.61471 


logc 


= 1.94549 


ft = 41.182 


c 


= 88.204 





26 



PLANE TRIGONOMETRY 







Case III 


A 


B 


Given A = 50° 2', 6 = 88 ; find B, a, c 
1. B = 90° -A= 39° 58'. 


V 


a 


2. ~ = tan A ; .'. a = b tan .4. 


/^oV 


G 


3. - = cos ^4. 

c 


6=88 
Fig. 17 


.'. b = c cos .4, and c = -■ 

cos ^4 



log a = log b + log tan .4 

log b = 1.94448 

log tan A = 10.07670 - 10 

loga = 2.02118 

a = 105.00 



logc = log &+ colog cos ^4 
log b = 1.94448 
cologcos,4 = 0.19223 
logc = 2.13671 
c = 137.00 




Case IY 



Given c = 58.40, a = 47.55; find A, 



B, b. 
1. sin A 



2. £ = 90° -A 



3. - = cot A ; .'. b = a cot yl . 



log sin ^4 = log a + colog c 

log a = 1.67715 

colog e = 8.23359 - 10 

log sin A = 9.91074-10 

A = 54° 31' 

B = 35° 29' 



log b = log a + log cot A 
log a = 1.67715 
log cot A = 9.85300 - 10 
log b = 1.53015 
b = 33.896 



THE RIGHT TRIANGLE 



27 



Given a = 40, b 



1. tan A — -• 



2. B = 90°-A. 



Case V 
find A, B, c. 



3. 



sm ^4. 



c sin ^4 , and c 



sin yl 




log tan A = log a + colog b 

loga= 1.60206 

colog b = 8.56864 - 10 

log tan 4 = 10.17070 - 10 

A == 55° 59' 

£ = 34° 1' 



log c = log #-h colog sin A 
log a = 1.60206 
colog sin ,1 = 0.08151 
logc =1.68357 
c = 48.258 



Note. In Cases IV and V the unknown side may also be found from 
the equations ' • 

(for Case IV) b = Vc 2 - a 2 = V(c + a)(c - a) ; 
(for Case V) c = Va 2 + b~. 

These equations express the values of b and c directly in terms of the 
two given sides ; and if the values of the sides are simple numbers (e.g., 5, 
12, 13), it is often easier to find b or c in this way. But this value of c is 
not adapted to logarithms, and this value of b is not so readily found by 
logarithms as the value of b given under Case IV. See also p. 23. 



SECTION XIV 
AREA OF THE RIGHT TRIANGLE 



The area of a triangle is equal to one-half the product of 
the base by the altitude ; therefore, if a and b denote the legs 
of a right triangle, and F the area, F = \ab. 

Hence ; the area may be found when a and b are known. 



28 



PLANE TRIGONOMETRY 



For example : Find the area, 
Case I (Sect. XIII, p. 25). 
A = 34° 28', c = 18.75. 
First find (as in Sect. XIII, 
p. 25) log a and log b. 
log F = log a + log b + colog 2 
log a = 1.02576 
log * = 1.18917 
colog 2 = 9.69897 - 10 
log F = 1.91390 
F = 82.016 



having given : 

Case IV (Sect. XIII, p. 26). 

a = 47.55, c = 58.40. 
First find (as in Sect. XIII, 
p. 26) log a and log b. 
log F = log a + log b + colog 2 
log a = 1.67715 
logb =1.53015 
colog 2 = 9.69897 - 10 
log F= 2.90627 
F = 805.88 



EXERCISE IX 



Solve the following triangles by logarithms, finding the 
angles to the nearest minute: 



1 


Given 




Required 




a=6, 


c=12. 


A =30°, 


£ = 60°, 


6=10.392. 


2 


.4=60°, 


6=4. 


B=30°, 


c=8, 


a=6.9282. 


3 


^4=30°, 


a = 3. 


E=00°, 


c=6, 


6 = 5.1961. 


4 


a=4, 


6=4. 


^=£=45°, 


c = 5.6568. 




5 


a=2, 


c= 2. 82843. 


^4=E=45°, 


6=2. 




6 


C = 627, 


4=23° 30'. 


£=66° 30', 


a =250. 02, 


6=575.0. 


7 


c=2280, 


^4=28° 5'. 


5=61° 55', 


a=1073.3, 


6=2011.5. 


8 


c=72.15, 


^4=39° 34'. 


£=50° 26', 


a=45.958, 


6=55.620. 


9 


c=l, 


4=36°. 


5=54°, 


a=0.58779, 


6=0.80902. 


10 


c = 200, 


5=21° 47'. 


^4 = 68° 13', 


a = 185.72, 


6=74.22. 


11 


c=93.4, 


J5=76°25'. 


.4 = 13° 35', 


a=21.936, 


6=90.788. 


12 


a=637, 


4 = 4° 35'. 


#=85° 25', 


6=7946, 


c = 7971.5. 


13 


a =48. 532, 


^4=36° 44'. 


5=53° 16', 


6=65.031, 


c = 81.144. 


14 


a=0.0008, 


^4 = 86°. 


£=4°, 


6=0.0000559 


c =0.000802. 


15 


6=50.937, 


J?=43°48'. 


^4=46° 12', 


a =53. 116, 


c=73.59. 


16 


6=2, 


B= 3° 38'. 


^4=86° 22', 


a=31.496, 


d=31.669. 



THE RIGHT TRIANGLE 



29 



17 


Gi 


VEN 


Required 


a = 992, 


5=76° 19'. 


4 = 13° 41', 


6 = 4074.5, 


c =4193. 5 


18 


a=73, 


£=68° 52'. 


A =21° 8', 


6=188.80, 


c = 202.47. 


19 


a=2.189, 


£=45° 25'. 


^4 = 44° 35', 


6 = 2.2211, 


c =3. 1185. 


20 


6-4, 


.4 =37° 56'. 


B =52° 4', 


a =3. 1176, 


c = 5.0714. 


21 


c = 8590, 


a=4476. 


4 =31° 24', 


JB=58°36 / , 


6 = 7332.8. 


22 


c = 86.53, 


0=71.78. 


^1 = 56° 3', 


5=33° 57', 


6 = 48.324. 


23 


c=9.35, 


a=8.49. 


^4=65° 14', 


£=24° 46', 


6 = 3.017. 


24 


c = 2194, 


6=1312.7. 


A = 53° 15', 


£=36° 45', 


a = 1758. 


25 


c = 30.69, 


6 = 18.256. 


4=53° 3(K, 


B =36° 30', 


a = 24.07. 


26 


a=38.313, 


6=19.522. 


.4=63°, 


£=27°, 


c = 43. 


27 


a = 1.2291, 


6 = 14.950. 


^4 = 4° 42', 


#=85° 18', 


c = 15. 


28 


a = 415.38, 


6=62.080. 


4=81? 30', 


5= 8° 30', 


c=420. 


29 


a = 13. 690, 


6 = 16.926. 


^1 = 38° 58', 


£=51° 2', 


c = 21".769. 


30 


c = 91.92, 


a=2.19. 


4= 1°22', 


J5=88°38', 


6 = 91.894 



Compute the unknown parts and also the area, having given : 

31. a = 5, b = 6. 36. c = 68, A = 69° 54'. 

32. a = 0.615, c = 70. 37. c = 27, B = 44° 4'. 

33. b = $2, c = VS. 38. a = 47, B = 48° 49'. 

34. a = 7, A = 18° 14'. 39. b = 9, £ = 34° 44'. 

35. J = 12, A = 29° 8'. 40. c = 8.462, B = 86° 4'. 

41. Find the value of F in terms of c and A. 

42. Find the value of F in terms of a and A 

43. Find the value of F in terms of b and .4. 

44. Find the value of F in terms of a and c. 

45. Given F = 58, a = 10 ; solve the triangle. 

46. Given F = 18, & = 5 ; solve the triangle. 

47. Given F= 12, ^ = 29°; solve the triangle. 

48. Given F = 100, c = 22 ; solve the triangle. 

49. Find the angles of a right triangle if the hypotenuse is 
equal to three times one of the legs. 



30 



PLANE TRIGONOMETRY 



50. Find the legs of a right triangle if the hypotenuse is 6, 
and one angle is twice the other. 

51. In a right triangle given c, and A = nB ; find a and b. 

52. In a right triangle the difference between the hypote- 
nuse and the greater leg is equal to the difference between 
the two legs. Find the angles. 




Fig. 20 

The angle of elevation of an object, or the angle of depres- 
sion, is the angle which a line from the eye to the object makes 
with a horizontal line in the same vertical plane. 

Thus, if the observer is at (Fig. 20), x is the angle of 
elevation of B, and y is the angle of depression of C. 

53. At a horizontal distance of 120 feet from the foot of a 
steeple, the angle of elevation of the top was found to be 
60° 30'. Find the height of the steeple. 

54. From the top of a rock that rises vertically 326 feet out 
of the water, the angle of depression of a boat was found to be 
24°. Find the distance of the boat from the foot of the rock. 

55. How far is a monument, in a level plain, from the eye, 
if the height of the monument is 200 feet and the angle of 
elevation of the top 3° 30' ? 

js^ 56. A distance AB is measured 96 feet along the bank of a 
river from a point A opposite a tree C on the other bank. 
The angle ABC is 21° 14'. Find the breadth of the river. 



THE RIGHT TRIANGLE 31 

57. What is the angle of elevation of an inclined plane if it 
rises 1 foot in a horizontal distance of 40 feet ? 

58. Find the angle of elevation of the sun when a tower 
120 feet high casts a horizontal shadow .70 feet long. 

59. How high is a tree that casts a horizontal shadow 80 feet 
in length when the angle of elevation of the sun is 50° ? 

60. A ship is sailing due northeast at a rate of 10 miles 
an hour. Find the rate at which she is moving due north, 
and also due east. 

61. In front of a window 20 feet high is a flower-bed 6 feet 
wide. How long is a ladder that will just reach from the 
edge of the bed to the window ? 

62. A ladder 40 feet long may be so placed that it will reach 
a window 33 feet high on one side of the street, and by turn- 
ing it over without moving its foot it will reach a window 21 
feet high on the other side. Find the breadth of the street. 

63. From the top of a hill the angles of depression of two 
successive milestones, on a straight level road leading to the 
hill, are observed to be 5° and 15°. Find the height of the hill. 

64. A fort stands on a horizontal plain. The angle of 
elevation at a certain point on the plain is 30°, and at a point 
100 feet nearer the fort it is 45°. How high is the fort ? 

65. From a certain point on the ground the angles of eleva- 
tion of the belfry of a church and of the top of the steeple 
were found to be 40° and 51°, respectively. From a point 300 
feet farther off, on a horizontal line, the angle of elevation of 
the top of the steeple is found to be 33° 45'. Find the distance 
from the belfry to the top of the steeple. 

66. The angle of elevation of the top C of an inaccessible 
fort observed from a point A is 12°. At a point B, 219 feet 
from A and on a line AB perpendicular to AC, the angle ABC 
is 61° 45'. Find the height of the fort. 



32 



PLANE TRIGONOMETRY 



SECTION XV 

THE ISOSCELES TRIANGLE 

An isosceles triangle is divided by the perpendicular from 
the vertex to the base into two equal right triangles. 

Therefore, an isosceles triangle is determined by any two 
parts that determine one of these right triangles. 

Let the parts of an isosceles tri- 
angle CAB (Fig. 21), among which 
the altitude CD is to be included, 
be denoted as follows : 

a = one of the equal sides, 

c = the base, 

h = the altitude, 

A = one of the equal angles, 

C = the angle at the vertex. 

For example : Given a and c ; required A, C, h. 

-i \ G c 

1. cos^4 = — = - — 
a Z a 

2. C + 2A -180°; .*. C = 180°-2,4 =2(90° -,4). 

3. h may be found by any one of the equations : 




= /»». 



whence 

Also, 
whence 



4 



/i = V(a + ic)(a-ic). 

— = 8m.A i and — =tan^4; 
a \e 

h = a sin A, and A = ictan^4. 



When c and h are known, the area can be found by the 
formula 



THE RIGHT TRIANGLE 33 

EXERCISE X 

Solve the following isosceles triangles, finding the angles 
to the nearest second : 



1. Given a and A 

2. Given a and C 

3. Given c and A 

4. Given c and C 

5. Given h and A 

6. Given h and C 

7. Given a and A 

8. Given c and A 



find C, c, h. 
find yl, c, h. 
find C, a, h. 
find J, a, A. 
find C, a, c. 
find A, a, c. 
find ^1, C, c. 
find A, C, a. 
9. Given a = 14.3, c = 11 ; find /l, C, A. 

10. Given a = 0.295, yl = 68° 10'; find c, h, F. 

11. Given c = 2.352, C = 69° 49'; find a, h, F. 

12. Given h = 7.4847, yl = 76° 14'; find a., c, 1\ 

13. Given a = 6.71, h = 6.6 ; find yl, C, c. 

14. Given c = 9, A = 20 ; find A, C, a. 

15. Given c = 147, F= 2572.5 ; find yl, C, a, h. 

16. Given h = 16.8, F = 43.68 ; find A, C, a, c. 

17. Find the value of F in terms of a and c. 

18. Find the value of F in terms of a and C 

19. Find the value of F in terms of a and A. 

20. Find the value of F in terms of h and C 

21. A barn is 40 x 80 feet, the pitch of the roof is 45°; 
find the length of the rafters and the area of the whole roof. 

22. In a unit circle what is the length of the chord corre- 
sponding to the angle 45° at the centre ? 

23. If the radius of a circle is 30, and the length of a 
chord is 44, find the angle subtended at the centre. 



34 



PLANE TRIGONOMETRY 



24. Find the radius of a circle if a chord whose length 
is 5 subtends at the centre an angle of 133°. 

25. What is the angle at the centre of a circle if the corre- 
sponding chord is equal to § of the radius ? 

26. Find the area of a circular sector if the radius of the 
circle is 12, and the angle of the sector is 30°. 

SECTION XVI 



THE REGULAR POLYGON 

Lines drawn from the centre of a regular polygon (Fig. 22) 
to the vertices are radii of the circumscribed circle ; and lines 
drawn from the centre to the middle points of the sides are 
radii of the inscribed circle. These lines divide the polygon 
into equal right triangles. Therefore, a regular polygon is 
determined by a right triangle whose sides are the radius of 
the circumscribed circle, the radius of the inscribed circle, and 
half of one side of the polygon. 

If the polygon has n sides, the angle of this right triangle 

. ' -p ^ " i ■ i * 1 (360°\ 180° 
at the centre of the polygon is equal to - 1 1 , or ; 

and the triangle may be solved when a side of the polygon or 

one of the radii is given. 

Let 
n = number of sides, 
c = length of one side, 
r = radius of circumscribed circle, 
h = radius of inscribed circle, 
p = the perimeter, 
F = the area. 

Then, by Geometry, 




Fig. 22 



THE RIGHT TRIANGLE 35 

EXERCISE XI 

Find the remaining parts of a regular polygon, given : 
*•" 1. 7i = 10, = 1. 3. n = 20, r = 20. 5. 7i = ll, F= 20. 
2. 7i = 18, r = 1. 4. 7i = 8, A = 1. 6. 7i = 7, F = 7. 
Find the side of : 

7. A regular decagon inscribed in a unit circle. 

8. A regular decagon circumscribed about a unit circle. 

9. If the side of an inscribed regular hexagon is 1, find the 
side of an inscribed regular dodecagon. 

10. Given n and c, and let b denote the side of the inscribed 
regular polygon having 2 n sides ; find b in terms of n and c. 

11. Compute the difference between the areas of a regular 
octagon and a regular nonagon if the perimeter of each is 16. 

12. Compute the difference between the perimeters of a regu- 
lar pentagon and a regular hexagon if the area of each is 12. 

Find the area of : 

13. The regular octagon formed by cutting away the corners 
of a square whose side is 1. 

14. A regular pentagon if its diagonals are each equal to 12. 

15. A regular polygon of 11 sides inscribed in a circle, if 
the area of an inscribed regular pentagon is 331.8. 

16. A circle inscribed in an equilateral triangle whose perim- 
eter is 20. 

17. A regular polygon of 15 sides inscribed in a circle, if 
the area of a regular inscribed polygon of 16 sides is 100. 

18. Find the perimeter of a regular dodecagon circum- 
scribed about a circle the circumference of which is 1. 

19. The area of a regular polygon of 25 sides is 40; find 
the area of the ring comprised between the circumferences of 
the inscribed and circumscribed circles. 



CHAPTER III 
GONIOMETRY 

SECTION XVII 
DEFINITION OF GONIOMETRY 

To prepare the way for the solution of the oblique triangle, 
we now proceed to extend the definitions of the trigonometric 
functions to angles of all magnitudes, and to deduce certain 
useful relations of the functions of different angles. 

That branch of Trigonometry which treats of trigono- 
metric functions in general, and of their relations, is called 
Goniometry. 

SECTION XVIII 

POSITIVE AND NEGATIVE QUANTITIES 

In measurements it is convenient to mark the distinction 
between two magnitudes that are measured in 
opposite directions, by calling one of them posi- 
tive and the other negative. 
X Thus, if OX (Fig. 23) is considered to be 
positive, then OX' is considered to be negative ; 
Y' and if OY is considered to be positive, then 

fig. 23 Qyi i s considered to be negative. 

When this distinction is applied to angles, an angle is 
considered to be positive, if the rotating line that describes 
it moves counter-clockwise, that is, in the direction opposite 

3G 



GONIOMETRY 37 

to the hands of a clock, and to be negative, if the rotating 
line moves clockwise, that is, in the same direction as the 
hands of a clock. ^__ B 

Arcs corresponding to positive angles are f /\ 

considered positive, and arcs corresponding to / A \ . 

negative angles are considered negative. \ ^v^i 

Thus, the angle A OB (Fig. 24) described by \^ J/&' 

a line rotating about from OA to OB is pos- 

. . , , Fig. 24 

itive, and the arc AB is positive; the angle 

A OB' described by the line rotating about from OA to OB' is 

negative, and the arc AB' is negative. 



SECTION XIX 

CO-ORDINATES OF A POINT IN A PLANE 

Let XX' (Fig. 25) be a horizontal line and let YY' be a line 
Y perpendicular to XX' at the point 0. 

Then the plane determined by the 
lines XX' and YY' is divided into 
four quadrants which are numbered 
^ I, II, III, IV. 

Any point in the plane is deter- 



II 



X' 

III 



mined by its distance and direction 
from each of the perpendiculars XX' 
and YY'. Its distance from YY', meas- 
ured on XX', is called the abscissa of 
the point; its distance from XX', measured on YY', is called 
the ordinate of the point. 

The abscissa and the ordinate of a point are called the 
co-ordinates of the point ; and the lines XX' and YY' are called 
the axes of co-ordinates. XX' is called the axis of abscissas or 
the axis of x ; YY' is called the axis of ordinates or the axis of y ; 
and the point is called the origin. 



38 



PLANE TRIGONOMETRY 



-VS 



B, 



A, 



£4 



B : 



A Z 

f 

Fig. 26 



In Fig. 26 the co-ordinates P lf P 2 , 
P 3 , P 4 are as follows : 

The abscissa of P x is OB x , 

and the ordinate of P 1 is C^; 
the abscissa of P 2 is O.B 2 , 

and the ordinate of P 2 is OA 2 ; 
, the abscissa of P 3 is 0.63, 

and the ordinate of P 3 is OA 3 ; 
the abscissa of P 4 is Oj5 4 , 

and the ordinate of P 4 is OA 4 . 



Abscissas to the WgrAtf of FF' are positive. 
Abscissas to the left of YY' are negative. 
Ordinates above XX' are positive. 
Ordinates below XX' are negative. 

Therefore, 
in Quadrant I, 

the abscissa is positive, the ordinate is positive ; 
in Quadrant II, 

the abscissa is negative, the ordinate is positive ; 
in Quadrant III, 

the abscissa is negative, the ordinate is negative ; 
in Quadrant IV, 

the abscissa is positive, the ordinate is negative. 



SECTION XX 

ANGLES OF ANY MAGNITUDE 

If the line OP (Figs. 27-30) is revolved about from OX 
as its initial position counter-clockwise, as shown by the curved 
arrows, the line during one revolution will form with OX all 
angles from 0° to 360°. 



GONIOMETRY 



39 



Any particular angle is said to be an angle of that quadrant 
in which its terminal side lies. 




Fig. 27 



Fig. 28 



Fig. 29 



Fig. 30 



Angles between 0° and 90° are angles of Quadrant I. 
Angles between 90° and 180° are angles of Quadrant II. 
Angles between 180° and 270° are angles of Quadrant III. 
Angles between 270° and 360° are angles of Quadrant IV. 




Fig. 31 



Fig. 32 



Fig. 33 



If the revolving line makes another revolution (Figs. 31-34). 
it will describe all angles from 360° to 720° ; and so on. 






II 


r \ 


v 

















-X IV / 


P X 


'in 






/ ii/^- 


""nLx 


\p 


1 [o 




1 


\ in^. 


-^IV 





Fig. 37 



Fig. 38 



If the line OP is revolved from OX clockwise (Figs. 35-38), 
it will describe all negative angles. 

Thus we arrive at the conception of an angle of any magni- 
tude, positive or negative. 



40 



PLANE TRIGONOMETRY 



SECTION XXI 



FUNCTIONS OF ANY ANGLE 



Figs. 39-42 show the functions in a unit circle drawn for 
an angle A OP in each quadrant, taken in order. The tangents 
to the circle are always drawn through A and B. 

Let the angle AOP formed with OA by the moving radius 
OP be denoted by x\ then, in each quadrant, 

sin x = MP, tan x = A T, sec x = OT, 

cos x = OM, cot x = BS, esc x = OS. 






B 






•\ J 


PyC 


AM ( 




A 









p 









B' 

Fig. 41 




GONIOMETRY 



41 



If the terminal line of any angle x extends through the 
vertex indefinitely both ways, and if the circumference of a 
unit circle cuts the terminal line at P, the axis of abscissas at 
A , and the axis of ordinates at B, then 

sin x = the ordinate of P ; 

cos x = the abscissa of P ; 

tan x == the tangent from A to meet the terminal line ; 

cot x = the tangent from B to meet the terminal line ; 

sec x = the segment of the terminal line between the vertex 

and the tangent ; 
esc x = the segment of the terminal line between the vertex 

and the cotangent. 

Sines and tangents extending from the axis of abscissas 
upwards are positive ; downwards, negative. 

Cosines and cotangents extending from the axis of ordinates 
towards the right are positive ; towards the left, negative. 

The signs of the secant and cosecant are determined by 
the signs of the cosine and sine, respectively. Therefore, 
secants and cosecants extending from the centre, in the direc- 
tion of the terminal line, are considered positive ; in the oppo- 
site direction, negative. Hence, 



Quadrant 


i 


11 


in 


IV 


sin and esc 
cos and sec 
tan and cot 


+ 
+ 
+ 


+ 


+ 


+ 



In Quadrant I all the functions are positive. 

In Quadrant II the sine and cosecant only are positive. 

In Quadrant III the tangent and cotangent only are positive. 

In Quadrant IV the cosine and secant only are positive. 



42 



PLANE TRIGONOMETRY 



The signs of all the functions of any quadrant are known 
when the signs of the sine and cosine are known. 

If the sine and cosine have like signs, the tangent and cotan- 
gent are positive ; if unlike signs, negative. The sine and 
cosecant have like signs ; the cosine and secant have like signs. 

SECTION XXII 



FUNCTIONS OF A VARIABLE ANGLE 

Let the angle A OP (Fig. 43) increase continuously from 0° 
to 360°. The values of its functions change as follows : 

1. The Sine. In the first quad- 
rant the sine MP increases from 
to 1 ; in the second it remains 
positive, and decreases from 1 to 
; in the third it is negative, and 
increases in absolute value from 
to 1 ; in the fourth it is nega- 
tive, and decreases in absolute 
value from 1 to 0. 

2. The Cosine. In the first 
quadrant the cosine OM decreases 
from 1 to ; in the second it be- 
comes negative, and increases in 
absolute value from to 1 ; in 
the third it is negative, and de- 
creases in absolute value from 1 

to ; in the fourth it is positive, and increases from to 1. 

3. The Tangent. In the first quadrant the tangent AT 
increases from tooo; in the second it becomes negative, 
and decreases in absolute value from oo to ; in the third it 
is positive, and increases from to oo ; in the fourth it is 
negative, and decreases in absolute value from oo to 0. 



5 
p') 


B 


s/ 




~-\ 






N / 


K 


1 M' 




/o 


M 


\ 




N' \ 


\p'" 


P" 








1 


r 





Fig. 43 



GONIOMETRY 



43 



4. The Cotangent. In the first quadrant, the cotangent 
BS decreases from oo to ; in the second it is negative, and 
increases in absolute value from to oo ; in the third and 
fourth quadrants, it has the same sign, and undergoes the 
same changes as in the first and second quadrants, respectively. 

5. The Secant. In the first quadrant, the secant OT 
increases from 1 to oo ; in the second it is negative, and 
decreases in absolute value from oo to 1 ; in the third it is 
negative, and increases in absolute value from 1 to oo ; in the 
fourth it is positive, and decreases from oo to 1. 

6. The Cosecant. In the first quadrant, the cosecant OS 
decreases from oo to 1 ; in the second it is positive, and 
increases from 1 to oo ; in the third it is negative, and 
decreases in absolute value from oo to 1 ; in the fourth it is 
negative, and increases in absolute value from 1 to oo. 

The limiting values of the functions are as follows : 



Sine 


0° 


90° 


180° 


270° 


360° 


±0 


+ 1 


±0 


-1 


±0 


Cosine 


+ 1 


±0 


-1 


±0 


+ 1 


Tangent 


±0 


±oo 


±0 


±00 


±0 


Cotangent 


±oo 


±0 


±00 


±0 


±oc 


Secant 


+ 1 


±oo 


-1 


±oo 


+ 1 


Cosecant 


±oo 


+ 1 


±00 


-1 


±O0 



Sines and cosines vary in value from +1 to — 1 ; tangents 
and cotangents, from + oo to — oo ; secants and cosecants, from 
+ oo to + 1, and from — 1 to — oo. 



In the table given above the double sign ± is placed before and oo . 
From the preceding investigation it appears that the functions always 
change sign in passing through and oo ; and the sign + or — prefixed 
to or oo simply shows the direction from which the value is reached. 



44 



PLANE TRIGONOMETRY 



SECTION XXIII 
FUNCTIONS OF ANGLES LARGER THAN 360° 

The functions of 360° + x are the same in sign and in 
absolute value as those of x ; for the moving radius has the 
same position in both cases. If n is a positive integer, 

The functions of (n x 360° + x) are the same as those of x. 

For example : The functions of 2200° (6 x 360° + 40°) are 
equal to the functions of 40°. 

SECTION XXIV 
EXTENSION OF FORMULAS 



The Formulas [1], [2], [3] established for acute angles on 
pp. 13, 14 hold true for all angles. Thus, in each quadrant 

MP 2 + OM 2 = OP 2 . 

Therefore, 

sin 2 x -f cos 2 x = 1. [1] 

We have in each quadrant from 
the similar triangles OMP, OAT, 
OBS the proportions 

A T : OA = MP : OM, 

or tan x : 1 = sin x : cos x ; 
MP : OP = OB: OS, 

or sin x : 1 = 1 : esc x ; 
OM: OP = OA: OT, 

or cos x : 1 = 1 : sec x ; 
AT:OA = OB : BS, 

or tan x : 1 = 1 : cot x. 




FIG. 44 



GONIOMETRY 45 



That is, tanx = ^ n -^. r 2 -| 

COS X L J 

sin x x csc x = 11 

cos x x sec x = 1 v • [3] 

tan x x cot x = 1 J 

Formulas [l]-[3] enable us, from a given value of one 
function, to find the absolute values of the other five func- 
tions, and also the sign of the reciprocal function. But in 
order to determine the proper signs to be placed before the 
other four functions, we must know the quadrant to which 
the angle in question belongs, or the sign of any one of these 
four functions ; for, by Sect. XXI, p. 40, it will be seen that the 
signs of any two functions that are not reciprocals determine 
the auadrant to which the angle belongs. 

Example. Given sin x = -f- f , and tan x negative ; find 
the values of the other functions. 

Since sin x is positive, x is an angle in Quadrant I or II ; 
but, since tan x is negative, Quadrant I is inadmissible. 



By [1], cosz=±Vl-^f = ±g. 



Since the angle is in Quadrant II, the minus sign must be 
taken, and we have 

cos x — — J . 
By [2] and [3], 

tan#=— f, cotx=— §, secx=— §, cscaj = |. 

EXERCISE XH 

1. Construct the functions of an angle in Quadrant II. 
What are their signs ? 

2. Construct the functions of an angle in Quadrant III. 
What are their signs ? 



46 PLANE TRIGONOMETRY 

3. Construct the functions of an angle in Quadrant IY. 
What are their signs ? 

4. What are the signs of the functions of the following 
angles : 340°, 239°, 145°, 400°, 700°, 1200°, 3800° ? 

5. How many angles less than 360° have the value of the 
sine equal to + f , and in what quadrants do they lie ? 

6. How many values less than 720° can the angle x have 
if cos x = -f- |-j and in what quadrants do they lie ? 

7. If we take into account only angles less than 180°, how 
many values can x have if sin x = f ? if cos x = i ? if cos x = 
-f? if tana; = |? ifcotcc=-7? 

8. Within what limits must the angle x lie if cos x = — § ? 
if cot x = 4 ? if sec x = 80 ? if esc a; = - 3 ? (If x < 360°.) 

9. In what quadrant does an angle lie if sine and cosine 
are both negative ? if cosine and tangent are both negative ? 
if cotangent is positive and sine negative ? 

10. Between 0° and 3600° how many angles are there whose 
sines have the absolute value f ? Of these sines how many 
are positive and how many negative ? 

11. In finding cos x by means of the equation cos x = 
± Vl — sin 2 £, when must we choose the positive sign and 
when the negative sign? 

12. Given cos x =— V^; find the other functions when x 
is an angle in Quadrant II. 

13. Given tan x = V3; find the other functions when x is 
an angle in Quadrant III. 

14. Given sec x = + 7, and tan x negative ; find the other 
functions of x. 

15. Given cot x = — 3 ; find all the possible values of the 
other functions. 

16. What functions of an angle of a triangle may be nega- 
tive ? In what case are they negative ? 



GONIOMETRY 



47 



17. Why may cot 360° be considered equal either to + oo 
or to — oo ? 

18. Obtain by means of Formulas [l]-[3] the other func- 
tions of the angles, given : 

(i) tan 90° = oo. (iii) cot 270° = 0. 

(ii) cos 180° = - 1. (iv) esc 360° = - ». 

19. Find the values of sin 450°, tan 540°, cos 630°, cot 720°, 
sin 810°, esc 900°. 

Compute the values of the following expressions : 

20. a sin 0° + b cos 90° - c tan 180°. 

21. a cos 90° - b tan 180° + c cot 90°. 

22. a sin 90° - b cos 360° + (a - b) cos 180°. 

23. (a 2 - b 2 ) cos 360° -±ab sin 270°. 

SECTIOX XXV 



REDUCTION OF FUNCTIONS TO THE FIRST QUADRANT 

In a unit circle (Fig. 45) draw two diameters PR and QS 
equally inclined to the horizontal diameter A A', or go that 
the angles A OP, A'OQ, A' OR, and 
AOS shall be equal. From the 
points P, Q, R, S let fall perpen- 
diculars to A A' i the four right 
triangles thus formed, with a 
common vertex at O, are equal ; 
because they have equal hypote- 
nuses (radii of the circle) and 
equal acute angles at O. There- 
fore, the perpendiculars PM, QX, 
RN, SM are equal, and are the 
sines of the angles AOP, AOQ, A OR, AOS, respectively. 




48 



PLANE TRIGONOMETRY 



Therefore, %n absolute value, 

sin A OP = sin A OQ = sin ^ 0# 



sin .4 05. 



And from Sect. XXIV, p. 44, it follows that in absolute 
value the cosines of these angles are also equal j and likewise 

the tangents, the cotangents, the 
secants, and the cosecants.* 

Hence, For every acute angle 
there is an angle in each of the 
higher quadrants whose functions, 
in absolute value, are equal to 
those of this acute angle. 

Let Z A OP = x, Z POB = y ; 
then x -f y = 90°, and the func- 
tions of x are equal to the co- 
named functions of y (Sect. V, p. 11) ; and 

Z AOQ (in Quadrant II) = 180° -x= 90° + y, 
ZAOR (in Quadrant III) = 180° + x = 270° - y, 
Z.AOS (in Quadrant IV) = 360° - x = 270° + y. 

Hence, prefixing the proper sign (Sect. XXI, p. 40), we have : 
Angle in Quadrant II 




sin (180° — x) = 


sinx. 


sin (90° + y) = cos y. 


cos (180° — x) = 


— cos X. 


cos (90° + y) = — sin y. 


tan (180° — x) = 


— tan x. 


tan (90° + y) = — cot y. 


cot (180° — x) = 


— cot X. 


cot (90° + y)= — tany. 



* In future, secants, cosecants, versed sines, and coversed sines will be 
disregarded. Secants and cosecants may be found by Formula [3], 
versed sines and coversed sines by VII and VIII, p. 5, if wanted, but 
they are seldom used in computations. 



G0N10METRY 49 

Angle in Quadrant III 

sin (180° -f x) = — sin x. sin (270° — y) — — cos y. 

cos (180° -f- ;z) = — cos x. cos (270° — y) = — sin y. 

tan (180° + x) = tan x. tan (270° — y) = cot y. 

cot (180° + x) = cot ac. cot (270° - y) = tan y. 

Angle in Quadrant IV 

sin (360° — a*) = — sin x. sin (270° + y) = — cos y. 

cos (360° — a;) = cos x. cos (270° + y) = sin y. 

tan (360° - x) = - tan x. tan (270° + y) = - cot y. 

cot (360° - a) = - cot x. cot.(270° + y) = - tan y. 

Remark. The tangents and cotangents may be found directly from 
the figure, or by Formula [2]. 

It is evident, from these formulas, 

1. The functions of all angles can be reduced to the functions 
of angles in the first quadrant, and therefore to functions of 
angles not greater than 45° (Sect. V, p. 11). 

2. If an acute angle is added to or subtracted from 180° or 
360°, the functions of the resulting angle are equal in absolute 
value to the like-named functions of the acute angle; but if an 
acute angle is added to or subtracted from 90° or 270°, the 
functions of the resulting angle are equal in absolute value to 
the co-named functions of the acute angle. 

3. A given value of a sine or cosecant determines two supple- 
mentary angles, one acute, the other obtuse; a given value of 
any other function determines only one angle: acute if the 
value is positive, obtuse if the value is negative. [See func- 
tions of (180° - x).~\ 



50 



PLANE TRIGONOMETRY 



SECTION XXVI 



FUNCTIONS OF ANGLES THAT DIFFER BY 90° 

The general form of two angles whose difference is 90° is x 
and 90° + x, and they must lie in adjoining quadrants. The 
relations between their functions were found in Sect. XXV, 
p. 48, but only for the case when x is acute. These relations, 
however, may be shown to hold true for all values of x. 

In a unit circle (Fig. 47) draw 
two diameters PR and QS per- 
pendicular to each other, and let 
fall to A A' the perpendiculars 
PM, QH, RK, and SN. The 
right triangles OMP, QHO, OKR, 
and SNO are equal, because they 
have equal hypotenuses and equal 
acute angles POM, OQH, ROK, and 




B' 
Fig. 47 


OSN. 


Therefore, 


OM= QH= OK = NS, 


d 


PM = OH= RK = ON. 



Hence, taking into account the algebraic sign, 
sin A OQ= cos A OP; sin A OS = cos A OR; 
cos A OQ = — sin A OP; cos AOS == — sin AOR ; 
sin A OR = cos A OQ ; sin (360° + A OP) = cos A OS ; 
cos AOR= - sin A OQ ; cos (360° + AOP) = - sin A OS. 

In all these equations, if x denotes the angle on the right- 
hand side, the angle on the left-hand side is 90° + x. 

Therefore, if x is an angle in any one of the four quadrants, 

sin (90° + x) = cos x, tan (90° -f- x) — — cot x. 

cos (90° + x) = — sin x, cot (90° + x) = — tan x. 



GOXIOMETRY 



51 



In like manner, it can be shown that all the formulas of 
Sect. XXV, p. 48, hold true, whatever the values of x and y. 

Hence, In every case the algebraic sign of the function 
of the resulting angle is the same as when x and y are both 
acute. 

SECTION XXVII 



FUNCTIONS OF A NEGATIVE ANGLE 

If the angle x is generated by the radius moving from the 
initial position OA to the terminal position OS, it will have the 
sign — , and its terminal side will 
be identical with its position for 
the angle 360° — x. Therefore, 
the functions of the angle — as 
are the same as those of the angle 
360° - x- or (Sect. XXV, p. 49), 

sin (— x) = — sin x, 
cos (— x) = cos x, 
tan (— a?) = — tan x, 
cot (— x) = — cot x. 




EXERCISE Xm 

1. Express sin 250° in terms of the functions of an acute 
angle less than 45°. 

Solution, sin 250° = sin (270° - 20°) = - cos 20°. 

Express the following functions in terms of the functions 
of angles less than 45° : 

2. sin 172°. 5. cot 91°. 8. sin 204°. 

3. cos 100°. 6. sec 110°. 9. cos 359°. 

4. tan 125°.. 7. esc 157°. . 10. tan 300°. 



52 PLANE TRIGONOMETRY 

11. cot 264°. 14. sin 163° 49'. 17. cot 139° 17'. 

12. sec 244°. 15. cos 195° 33'. 18. sec 299° 45'. 

13. esc 271°. 16. tan 269° 15'. 19. esc 92° 25'. 

Express all the functions of the following negative angles 
in terms of the functions of positive angles less than 45° : 

20. - 75°. 22. - 200°. 24. - 52° 37'. 

21. - 127°. 23. - 345°. 25. - 196° 54'. 

26. Find the functions of 120°. 

Hint. 120° = 180° - 60°, or 90° + 30° ; then apply Sect. XXV, p. 48. 
Find the functions of the following angles : 

27. 135°. 29. 210°. 31. 240°. 33. -30°. 

28. 150°. 30. 225°. 32. 300°. 34. -225°. 

35. Given sin x = — ^ V2, and cos x negative; find the other 
functions of x, and the value of x. 

36. Given cot x = — V3, and x in Quadrant II ; find the 
other functions of x, and the value of x. 

37. Find the functions of 3540°. 

38. What angles less than 360° have a sine equal to — \ ? 
a tangent equal to — V3 ? 

39. Which of the angles mentioned in Examples 27-34 
have a cosine equal to — £ V2 ? a cotangent equal to — V3 ? 

40. What values of x between 0° and 720° will satisfy the 
equation sin x = + \ ? 

41. Eind the other angle between 0° and 360° for which the 
corresponding function (sign included) has the same value as 
sin 12°, cos 26°, tan 45°, cot 72°, sin 191°, cos 120°, tan 244°, 
cot 357°. 



GOXIOMETRY 



53 



42. Given tan 238° = 1.6: find sin 122 c . 

43. Given cos 333° = find tan 117 3 . 

Simplify the following expressions: 

44. a cos (90° - *) + b cos | 90 € - 

45. m cos « 90° - x) sin (90° - x). 

46. (a - b) tan (90" - x ~-(a + b) cot (90° - 

47. B* + ft*-2«fl cos 180" - 

48. sin (90° + x) sin (180° + x) + cos (90° + x) cos (180° - x). 

49. cos(l80°+^cos,27n 5 -^)_sin(180°-|-jrjsin(2;" ; _ 

50. tan x + tan (— y) - tan (1m ,: - 

51. Foi what values of x is the expression sin x -4- c 
positive, and for what values negative ? 

52. Answer the questions of Example 51 for sin x — cos x. 
5 3. Find the functions of x — 90° in functions of x. 

54. Find the functions of x — 180° in functions of x. 

SECTION XXVIII 
FUNCTIONS OF THE SUM OF TWO ANGLES 



In a unit circle | Fig. 49) let the angle A OB = x. the angle 
= y; then the angle AOC = 

x + y. 

In order to express sin (x 4- y) 
and cos (x -f y) in terms of the 
sines and cosines of x and y. draw 
CF J_ OA. CD _L 05. DE _ 0.1. 
Z>G _L CF: then CD = sin y, 0D = 
cos ?/, and the angle DCG = x. 
Also, u F E 

sin(x + y) = Ci 7 = BE — CO. Fig. 49 




54 



PLANE TRIGONOMETRY 



7") 77 

Now = sin x ; hence, DE = sin x X OD = sin x cos y. 

And — — = cos x ; hence, 

CG = cos cc X CD = cos # sin y. 
Therefore, 
sin(x + y) = sinxcosy + cosxsiny. [4] 

Again, 

cos (x + 3/) = OF = OE — DG. 
OE 
OD 
OE = cos x X OD = cos x cos y. 




cos x ; hence, 



DG . , . . 

——■ = sin a; ; hence, DG = sin as X CD = sin z sin y. 

CD ' ' ^ 



Therefore, 



cos (x + y) = cos x cos y — sin x sin y. [5] 




In this proof x and y, and also the sum x + y, are assumed 
to be acute angles. If the sum 
x + y of the acute angles # and 
y is obtuse, as in Fig. 51, the 
proof remains, word for word, 
the same as above, the only dif- 
ference being that the sign of 
OF will be negative, as DG is 
now greater than OE. The above formulas, therefore, hold 
true for all acute angles x and y. 

If these formulas hold true for any two acute angles x and 
y, they hold true when one of the angles is increased by 90°. 
Thus, if for x we write x ( = 90° + x, then, by Sect. XXV, 
p. 48, 

sin (x' + y) = sin (90° + x + y) = cos (as + y), 
cos (x' + y) = cos (90° + # + 3/) = — sin (x + y)- 



GOXIOMETRY 55 

Hence, by [5], sin (x' -f y) = cos # cos ?/ — sin x sin ?/, 
by [4], cos (x 1 + ?/) = — sin x cos y — cos x sin y. 

Now, by Sect. XXV, p. 48, 

cos x = sin (90° + x) — sin £', 
sin x = — cos (90° + jc) = — cos x'. 

Substitute these values of cos x and sin x, then 

sin (V + y) — sm iC ' cos 2/ + cos x ' sm 2/> 
cos (V + 2/) — cos a:' cos y — sin x' sin y. 

It follows that Formulas [4] and [5] hold true if either 
angle is repeatedly increased by 90° ; therefore they apply to 
all angles whatever. 

By Sect. XXIV, p. 45, Formula [2], 

sin (x + y) sin x cos y + cos x sin ?/ 

tan (x + y) = ; *f = ! — : : — - • 

x y cos (x + y) cos x cos ?/ — sin x sin ^/ 

If we divide each term of the numerator and denominator 
of the last fraction by cos x cos y, we have 



tan (x + y) = 



sin x sm y 

+ 

cos x cos y 

sin x sin y 

COS X cos ?/ 



That is, tan(x + y) = tanx + tan y . [6] 

v ^ Jy 1 -tanxtany L J 

cos (a; -f- ?/) cos x cos ?/ — sin x sin j/ 

Also, cot (x + y)= . ) •'[ =-. ^« 

x y sm (x -f- y) sin # cos ?/ — cos x sm ^/ 

I>ivide each term of the numerator and denominator by 

i • ^ i. cos x j cos V *. 

sm a* sm y, rememberinsr that — = cot x and — — - = cot y; 

. " 5 sm * sm y J 

we nave 

cot x cot y — 1 r __, 

cot (x + y) = • ["] 

v ^ J; coty + cotx L J 



56 



PLANE TRIGONOMETRY 



SECTION XXIX 



FUNCTIONS OF THE DIFFERENCE OF TWO ANGLES 

In a unit circle (Fig. 52) let the angle AOB = x, COB = y ; 
then the angle AOC — x — y. 

In order to express sin (x — y) 
and cos {x — y) in terms of the 
sines and cosines of x and y, draw 
CF _L OA, CD _1_ OB, DE _L 0^4, 
DG ± FC prolonged ; then CD = 
sin y, OB = cos y, and the angle 
DCG = x. 

Now sin (x-y) = CF= DE-CG. 

DE 

— — = sin ic j hence, DE = sin cc x OD = sin x cos y. 




CG 



CD 

Therefore, 
Again, 

OE 
OD 
DG 
CD 

Therefore, 



= cos x ; hence, CG — cos x x CD — cos x sin y. 



sin (x — y) = sin x cos y — cos x sin y. 

cos (x — y) = OF = OE + DG. 

— ^ = cos x ; hence, OE = cos ;c x 0D = cos ic cos y. 



[8] 



= sm# 



hence, DG = sin x x CD = sin x sin y. 

cos (x — y) = cos x cos y + sin x sin y. [9] 

In this proof, both x and y are assumed to be acute angles ; 
but, whatever be the values of x and y, the same method of 
proof will always lead to Formulas [8] and [9], when due 
regard is paid to the algebraic signs. 

The general application of these formulas may be at once 
shown by deducing them from the general formulas estab- 
lished in Sect. XXVIII, p. 54, as follows: 



GONIOMETRY 57 

It is obvious that (x — y) -f y = x. If we apply Formulas 

[4] and [5] to (x — y) + y, then 

sin J (a; — y) + ?/ j or sin x = sin (x — y) cos y + cos (x — ?/) sin y, 

cos j (ic — y) + y J or cosz = cos (x — y) cosy — sin (x — y) sin y. 
Multiply the first equation by cos y, the second by sin y, 
sin x cos y = sin (x — y) cos 2 ?/ + cos (x — y) sin y cos ?/, 
cos x sin y =— sin (# — ?/) sin 2 ?/ -f cos (x — y) sin y cos ?/ ; 

whence, by subtraction, 

sin x cos y — cos # sin y = sin (# — y) (sin 2 y + cos 2 ?/). 
But sin 2 ?/ + cos 2 ?/ = 1 (Sect. XXIV, p. 44). 
Therefore, by substitution and transposition, 

sin (x — y) = sin x cos y — cos x sin y. 

Again, if we multiply the first equation by sin y, the second 
equation by cos y, and add the results, we obtain, by reducing, 
cos (x — y) = cos x cos y + sin x sin y. 

Therefore, Formulas [8] and [9], like [4] and [5], from 

which they have been derived, are universally true. 

From [8] and [9], by proceeding as in Sect. XXVIII, p. 55, 

we obtain 

. v tan x — tan y 

tan(x-y) = £_. [101 

1 + tan x tan y L J 

... / \ cot x cot y + 1 

cot(x-y) = £_L_. fill 

v cot y — cot x L J 

Formulas [4]-[ll] may be combined as follows: 

sin (x ± y) = sin x cos y ± cos x sin y, 

cos (x ± y) = cos x cos y qz sin x sin y, 

, . . tan x ± tan ?/ 

tan (x±y) = - — — > 

v y 1 op tan x tan y 

, , , s cot # cot y n: 1 
cot (x ± y) = — - — , ^ J — 
x ' cot?/±cot;e 



58 PLANE TRIGONOMETRY 

SECTION XXX 
FUNCTIONS OF TWICE AN ANGLE 



If y = 


= x, Formulas [4]-[7] become 




sin 2 x = 2 sin x cos x. 




cos 2 x = cos 2 x — sin 2 x. 




2 tan x 
tan 2 x — 

1 — tan 2 x 




cot 2 x — 1 

cot 2 x — 



2 cot X 



[12] 
[13] 

[14] 
[15] 



By these formulas the functions of twice an angle may be 
found when the functions of the angle are given. 

SECTION XXXI 

FUNCTIONS OF HALF AN ANGLE 

Formula [1] is cos 2 x + sin 2 z = 1. 

Formula [13] is cos 2 a: — sin 2 ic = cos 2x. 

Subtract, 2 sin 2 a; = 1 — cos 2 x. 

Add, 2 cos 2 x = 1 + cos 2x. 



Whence, 
sin x 



\l — cos 2 x . /l + cos 2 x 
±\ g ' G0SX==± \ 2 

These values, if z is put for 2 x, and hence ■£- z for x, become 

*t.=±<jL=J£t [i6] 



i _i_ /l + cos z _„___ 

cos|z=±^ 2 • [17] 



GONIOMETRY 59 

Hence, by division (Sect. XXIV, p. 45), 

tan^=±Ji^iI. [18] 

» 1 + cos z L J 



c tiz=±yji±^l. [19] 

* 1 — COS z L J 

By these formulas the functions of half an angle may be 
computed when the cosine of the entire angle is given. 

The proper sign to be placed before the root in each case 
depends on the quadrant in which the angle £ z lies (Sect. 
XXII, p. 42). 

Let the student show from Formula [18] that 



tan i B = \/- -■ (See p. 23, Note.) 



SECTION XXXII 

SUMS AND DIFFERENCES OF FUNCTIONS 

From [4], [5], [8], and [9], by addition and subtraction, 
sin (x + y) + sin (x — y) = 2 sin x cos y, 
sin (x 4- y) — sin (x — y)= 2 cos x sin y, 
cos (x 4- y) 4 cos (x — y) = 2 cos x cos y, 
cos (a; + y) — cos (a: — y) = — 2 sin x sin y ; 

or, by making x + y = A, and x — y = B, . and, therefore, 
a; = i(A + 5), and y = i(A - 5), 

sin A 4 sin B = 2 sin £ (A + B) cos -J- (A — B). [20] 

sin A — sin B = 2 cos % (A 4- B) sin J (A — B). [21] 

cos A 4 cos B = 2 cos i (A 4 B) cos \ (A — B). [22] 

cos A — cos B = — 2 sin J- (A 4- B) sin £ (A — B). [23] 



60 PLANE TRIGONOMETRY 

From [20] and [21], by division, we obtain 

sin A + sin B , .' 

- — — ! — : — - = tan4-(^ +B)cotl(A -B): 

or, since cot \ (A — B) = 



tan ±(A — B) 

sin A + sin B _ tan \ (A + B) 
sin A — sin B ~~ tan \ (A — B) 

EXERCISE XIV 



[24] 



1. Find the value of sin (x + y) and cos (# -f ?/) when 
sin z = |, cos a; = f , sin ?/ = ^, cos y = If. 

2. Find sin (90° - y) and cos (90° - y) by making x = 90° 
in Formulas [8] and [9]. 

Find, by Formulas [4]-[ll], the first four functions of: 

3. 90° + y. 8. 360° - y. 13. - y. 

4. 180°-?/. 9. 360° + ?/. 14. 4:5° -y. 

5. 180° + y. 10. a; -90°. 15. 45° + ?/. 

6. 270° -y. 11. a: -180°. 16. 30° + ?/. 

7. 270° + y. 12. a; - 270°. 17. 60° - y. 

18. Find sin 3 a: in terms of sin x. 

19. Find cos 3 a; in terms of cos x. 

20. Given tan -J- x = 1 ; find cos x. 

21. Given cot -J- x = V3 ; find sin x. 

22. Given sin x = 0.2 ; find sin ^ x and cos -J a;. 

23. Given cos x = 0.5; find cos 2 a; and tan 2 as. 

24. Given tan 45° = 1 ; find the functions of 22° 30'. 

25. Given sin 30° = 0.5 ; find the functions of 15°. 

sin 33° + sin 3° 



26. Prove that tan 18° = 



cos 33° + cos 3 C 



GONIOMETKY 61 



Prove the following formulas : 

• o 2 tan x nrx , sin x 

27. sin2cc = - — - — - — 29. tan^a; 



1 + tan 2 x 1 -f- cos x 

n 1 — tan 2 ^ _ , , sin a; 

28. cos2x = - — — • 30. cot£a; = - 

1 + tan 2 # 1 — cos x 



31. sin -J- aj ± cos %x = V 1 ± sin a;. 

tan a: ± tan ?/ , , 

32. ; = ± tan x tan y. 

cot x ± cot y 

,*ro x 1 — tan x 

33. tan (45° — a;) = 

v 1 + tan x 

If A, B, C are the angles of a triangle, prove that : 

34. sin A + sin B + sin (7 = 4 cos ^- y1 cos ^ B cos £ C. 

35. cos A + cos B + cos (7 = 1 + 4 sin -J- A sin -J- i? sin £ C. 

36. tan ^4 + tan B + tan C = tan .1 x tan B x tan C. 

37. cot^A + cot £ J3 + cot £ C = cot £ .4 x cot££x cot^C. 

Change to a form more convenient for logarithmic compu- 
tation : 

38. cot x + tan x. 43. 1 -f- tan x tan y. 

39. cot x — tan x. 44. 1 — tan x tan y. 

40. cot x -f tan ?/. 45. cot a: cot ?/ + 1. 

41. cot x — tan ?/. 46. cot x cot y — 1. 

1 — cos 2 a; tan x -f tan ?/ 

42. — 47. -« 

1 -f cos 2 x - cot x + cot y 

SECTION" XXXIII 

ANTI-TRIGONOMETRIC FUNCTIONS 

If i/ is any trigonometric function of an angle x, then ae is 
said to be the corresponding anti-trigonometric function of y. 
Thus, if y = sin x, x is the anti-sine or inverse sine of 3/. 



62 PLANE TRIGONOMETRY 

The anti-trigonometric functions of y are written 
sin -1 ?/, tan -1 y, sec -1 ?/, vers -1 ?/, 

cos -1 ?/, cot -1 ?/, esc -1 ?/, covers -1 ?/. 

These are read, the angle whose sine is y, and so on. 
For example, sin 30° == \ ; hence, 30° = sin -1 !-. Similarly. 
90° = cos -1 = sin -1 l, and 45° = tan -1 l = sin -1 | V2, etc. 
The symbol _1 must not be confused with the exponent — 1. Thus, 

sin— 1 x is a very different expression from , which would be written 

sin x 
(sin x)— 1 . On the continent of Europe mathematical writers employ the 
notation arc sin, arc cos, etc., for sin- 1 , cos- 1 , etc. 

There is an important difference between the trigonometric 
and the anti-trigonometric functions. When an angle is given, 
its functions are all completely determined ; but when one of 
the functions is given, the angle may have any one of an indefi- 
nite number of values. Thus, if sin y = -J-, y may be 30°, or 
150°, or either of these increased or diminished by any integral 
multiple of 360° or 2 it, but cannot take any other values. 
Accordingly, sin -1 -J- = 30° ± 2 nir, or 150° ± 2 nir, where n is 
any positive integer. Similarly, tan - 1 1 = 45° ± 2 nir or 
225° ± 2 nir ; i.e., tan -1 l = 45° ± nir. 

Since one of the angles whose sine is x and one of the angles 
whose cosine is x together make 90°, and since similar rela- 
tions hold for the tangent and cotangent, for the secant and 
cosecant, and for the versed sine and coversed sine, we have 

i i -i 7r 
sec -1 ic -f esc x x =—> 

vers -1 a; -f- covers - x x — — t 

L 

where it must be understood that each equation is true only 
for a particular choice of the various possible values of the 
functions. Thus, if x is positive, and if the angles are always 
taken in the first quadrant, the equations are correct. 



sin" 


- x x 


+ cos - 


' X X 


= 


IT 

2' 


tan" 


~ x x 


H-cot - 


-*x 


= 


7T 

2* 



GONIOMETRY 63 

EXERCISE XV 

1. Find all the values of the following functions : 

sin" 1 $ V5, tan" 1 J V3, vers -1 £, cos" 1 (- £ V2), 
csc _1 V2, tan -1 cc, sec -1 2, cos -1 (— i V3). 

2. Prove that 

sin -1 (— x) = — sin -1 # ; cos -1 (— x) = it — cos -1 a\ 

3. If sin -1 a; + sin -1 ?/ = ir, prove that x = y. 

4. If y = sin -1 -J, find tan y. 



5. Prove that eos(sin -1 a:) = Vl — x 2 . 

6. Prove that cos (2 sin -1 x) = 1 — 2 x\ 

x I ?/ 

7. Prove that tan (tan -1 a: -f- tan -1 y) — — • 

v 7 1 — xy 

8. If x = VJ, find all the values of sin -1 a- -f cos -1 a. 

9. Prove that tan -1 ( — , ) = sin -1 a:. 

VVl -x 2 J 

10. Pind the value of sin (tan -1 -^). 

11. Pind the value of cot (2 sin -1 1). 

12. Pind the value of sin (tan -1 ^- + tan -1 -J). 

13. If sin -1 a; = 2 cos -1 a-, find x. 

2x 

14. Prove that tan (2 tan -1 a-) = - — — - • 

y 1 — x- 

2x 

15. Prove that sin (2 tan -1 a-) = 5 - 

v J 1 + x 2 



CHAPTER IV 

THE OBLIQUE TRIANGLE 

SECTION XXXIV 

LAW OF SINES 

Let A, B, C denote the angles of a triangle ABC (Figs. 53 
and 54), and a, b, c, respectively, the lengths of the opposite 
sides. 

Draw CD A.AB, and meeting AB (Fig. 53) or AB produced 
(Fig. 54) at D. Let CD = h. 

C 
C 





In either figure, - = sin A. 



In Fig. 53, 
In Fig. 54, 



- = sin B. 
a 



= sin(180°-5)=sm5. 



Therefore, whether h lies within or without the triangle, we 
obtain, by division, 

l = ~ [25] 

b sinB L J 

64 



THE OBLIQUE TRIANGLE 



65 



By drawing perpendiculars from the vertices A and B to 
the opposite sides, we may obtain, in the same way, 

b 

C 



sin B 



sin A 



sin C 



sin C 
Hence the Law of Sines : 

The sides of a triangle are proportional to the sines of the 
opposite angles. 

If we regard these three equations as proportions, and take 
them by alternation, it is evident that they may be written in 
the symmetrical form 

a b c 

sin A sin B sin C 

Each of these equal ratios has a simple geometrical meaning 
which will appear if the LaAv of Sines is proved as follows : 

Circumscribe a circle about the 
triangle ABC (Fig. 55), and draw 
the radii OB, OC. Let it denote 
the radius. Draw OM _L BC. By 
Geometry, the angle BOC = 2 A ; 
hence, the angle BOM = A, then 
BM = R sin BOM = R sin A. 

.'. BC or a = 2 R sin A. 
In like manner, 
b = 2 R sin B, and c = 2 R sin C. 
Whence we obtain 

a b 




Fig. 55 



2R = 



c 



sin A sin B sin C 



That is : The ratio of any side of a triangle to the sine of 
the opposite angle is numerically equal to the diameter of the 
circumscribed circle. 



66 



PLANE TRIGONOMETRY 



SECTION XXXV 



LAW OF COSINES 



This law gives the value of one side of a triangle in terms 
of the other two sides and the angle included between them. 





a 2 = h 2 + BD- 
BD = c — AD. 
BD = AD — c; 
BD 2 = AD 2 -2cxAD + c 2 . 

h 2 + AD 2 + c 2 -2cxAD. 



h 2 + AD 2 = b 2 , 



In Tigs. 56 and 57, 
In Fig. 56, 
In Fig. 51, 
In either case, 
Therefore, in all cases, 
Now, 
and AD = b cos A. 

Therefore, a 2 = b 2 + c 2 - 2 be cos A. [26~\ 

In like manner it may be proved that 

b 2 = a 2 + c 2 — 2 ac cos B, 
■ c 2 = a 2 + b 2 -2ab cos C. 

The three formulas have precisely the same form, and the 
Law of Cosines may be stated as follows : 

The square of any side of a trianyle is equal to the sum of 
the squares of the other two sides diminished by twice their 
product into the cosine of the included angle. 



THE OBLIQUE TRIANGLE 67 

SECTION XXXVI 
LAW OF TANGENTS 



By Sect. XXXIV, p. 64, a : b = sin A : sin B ; 
whence, by the Theory of Proportion, 

a — b sin A — sin B 



a + b sin A + sin B 




But by [24], p. 60, 




sin A — sin B tan \ (A — B) 




sin A + sin B tan £ (J. +5) 




Therefore, '"N^tffS- 
a + b tan £ (A + B) 




By merely changing the letters, 




a — c tan \ (A — C) b — c tan $ (B - 


-c) 



[27] 



a + c tan 1 (.4 + C) 6 + c tan £ (5 + C) 

Hence the Law of Tangents : 

The difference of two sides of a triangle is to their sum as 
the tangent of half the difference of the opposite angles is to 
the tangent of half their sum. 

Note. If in [27] b > a, then B>A. The formula is still true, but to 
avoid negative numbers the formula in this case should be written 
b — a _ tan \ (B — A) 
b + a ~ tan £ (B + A) 

EXERCISE XVI 

1. What do the formulas of Sect. XXXIV, p. 64, become 
when one of the angles is a right angle ? 

2. Prove by means of the Law of Sines that the bisector 
of an angle of a triangle divides the opposite side into parts 
proportional to the adjacent sides. 



68 PLANE TRIGONOMETRY 

3. What does Formula [26] become when A = 90° ? when 
J. = 0° ? when A = 180° ? What does the triangle become in 
each of these cases ? 

Note. The case when A = 90° explains why the theorem of Sect. 
XXXV, p. 6Q, is sometimes called the Generalized Theorem of Pythagoras. 

4. Prove (Figs. 56 and 57) that whether the angle B is 
acute or obtuse c = a cos B + b cos A. What are the two sym- 
metrical formulas obtained by changing the letters ? What 
does the formula become when B = 90° ? 

5. From the three following equations (found in the last 
example) prove the theorem of Sect. XXXV, p. 66 : 

c =z a cos B + b cos A, 
b = a cos C + c cos A, 
a = b cos C + c cos B. 

Hint. Multiply the first equation by c, the second by &, the third 
by a ; then from the first subtract the sum of the second and third. 

6. In Formula [27] what is the maximum value of -J- (A —B) ? 

7. Find the form to which Formula [27] reduces, and 
describe the nature of the triangle, when 

(i) C = 90° ; (ii) A - B = 90°, and B = C. 

SECTION XXXVII 

THE GIVEN PARTS 

The formulas established in Sects. XXXIV-XXXVI, pp. 

64-67, together with the equation 
A + B + C = 180°, are sufficient 
for solving every case of an 
oblique triangle. The three parts 
that determine an oblique triangle 
may be : 




THE OBLIQUE TRIANGLE 



69 



I. One side and two angles ; 

II. Two sides and the angle opposite one of these sides ; 

III. Two sides and the included angle ; 

IV. The three sides. 

SECTION XXXVIII 
SOLUTION OF AN OBLIQUE TRIANGLE 

Case I 

Given one side a and two 
angles A and B; find the re- 
maining parts C, b, and c. 

1. C = 1S0°-(A + B). 

2 h = sinB ; • i = 

a sin A 

e sinC. 

sin A sin A sin A 




3. - = 



A C 


j 


Fig. 


59 


a sin B a 


X sin B. 


sin A sin A 


a sin C a 


X sin C. 



Example, a = 24.31, A = 45° 18', B = 22° 11'. 
The work may be arranged as follows : 



a = 24.31 
A -45° 18' 
B = 22° 11' 



,4 +£ = 67° 29' 
C = 112°31 



log a = 1.38578 

colog sin A = 0.14825 

log sin B = 9.57700 

log fi = 1.11103 

b = 12.913 



= 1.38578 

= 0.14825 

log sin C = 9.96556 

log c = 1.49959 

c = 31.593 



Note. When — 10 is omitted after a logarithm or cologarithm, it 
must be remembered that the log or colog is 10 too large. 

EXERCISE XVII 



2. 



Given a = 500, 
find C = 123° 12', 

Given a = 795, 
find C = 55° 20', 



A = 10° 12', 
b = 2051.5, 

A = 79° 59', 
h = 567.69, 



B = 46° 36' 
c = 2362.6. 

B = 44° 41' 
c = 663.99. 



70 PLANE TRIGONOMETRY 



3. 


Given a = 804, 


A = 99° 55', 


£ = 45°1'; 




find C = 35° 4', 


& = 577.31, 


c = 468.93. 


4. 


Given a = 820, 


4 = 12° 49', 


5 = 141° 59'; 




find C = 25° 12', 


b = 2276.6, 


c = 1573.9. 


5. 


Given c = 1005, 


A = 78° 19', 


B = 54:°2T; 




find C = 47° 14', 


a = 1340.6, 


b = 1113.8. 


6. 


Given 6 = 13.57, 


5 = 13° 57', 


C = 57°13'; 




find A = 108° 50', 


a = 53.276, 


c = 47.324. 


7. 


Given a = 6412, 


4 = 70° 55', 


C = 52°9'; 




find B = 56° 56', 


5 = 5685.9, 


c = 5357.5. 


8. 


Given b = 999, 


A = 37° 58', 


C = 65°2'; 




find 5 = 77°, 


a = 630.77, 


c = 929.48. 



9. In order to determine the distance of a hostile fort A 
from a place i?, a- line BC and the angles ABC and 2?C\4 were 
measured and found to be 322.55 yards, 60° 34', and 56° 10', 
respectively. Find the distance AB. 

10. The angles B and C of a triangle ABC are 50° 30' and 
122° 9', respectively, and BC is 9 miles. Find AB and AC. 

11. Two observers 5 miles apart on a plain, and facing each 
other, find that the angles of elevation of a balloon in the 
same vertical plane with themselves are 55° and 58°, respec- 
tively. Find the distance from the balloon to each observer, 
and also the height of the balloon above the plain. 

12. In a parallelogram given a diagonal d and the angles 
x and y which this diagonal makes with the sides ; find the 
sides. Find the sides if d = 11.237, cc = 19° 1', and y = 42° 54'. 

13. A lighthouse was observed from a ship to bear N. 34° E. ; 
after the ship sailed due south 3 miles it bore N. 23° E. Find 
the distance from the lighthouse to the ship in each position. 

Note. The phrase to bear N. 34° E. means that the line of sight to 
the lighthouse is in the northeast quarter of the horizon and. makes, 
with a line due north, an angle of 34°. 



THE OBLIQUE TRIANGLE 71 

14. In a trapezoid given the parallel sides a and b, and the 
angles x and y at the ends of one of the parallel sides ; find 
the non-parallel sides. Compute the results when a = 15, 
b = 7, x = 70°, y = 40°. 

Solve the following examples without using logarithms : 

15. Given b = 7.07107, A = 30°, C = 105° ; find a and c. 

16. Given c = 9.562, A = 45°, B = 60° ; find a and b. 

17. The base of a triangle is 600 feet and the angles at the 
base are 30° and 120°. Find the other sides and the altitude. 

18. Two angles of a triangle are, the one 20°, the other 40°. 
Find the ratio of the opposite sides. 

19. The angles of a triangle are as 5 : 10 : 21, and the side 
opposite the smallest angle is 3. Find the other sides. 

20. Given one side of a triangle equal to 27, the adjacent 
angles equal each to 30° ; find the radius of the circumscribed 
circle. (See Sect. XXXIV, p. 65.) 

SECTION XXXIX 

Case II 

Given two sides a and b and the angle A opposite the 
side a ; find the remaining parts B, C, c. 

This case, like the preceding case, is solved by means of 
the Law of Sines. 

«. sin.B ^ 6 sin .4. 

Since — = -j therefore sin B = ■ > 

sm A a a 

C = 180° -(A + B). 

. n . c sin C . a sin C 

And since - = — > therefore c = —. 

a sin A sin A 



72 



PLANE TRIGONOMETRY 




Pig. 60 



When an angle is determined by its sine it admits of two 
values which are supplements of each other (Sect. XXV, 
p. 48) ; hence, either value of B may be taken unless excluded 
by the conditions of the problem. 

If a>b, then by Geometry A > B, and B must be acute 

whatever be the value of A ; for 
a triangle can have only one obtuse 
angle. Hence, there is one, and 
only one, triangle that will satisfy 
the given conditions. 

If a = b, then by Geometry 
A = B ; both A and B must be 
acute, and the required triangle is 
isosceles. 
If a < b, then by Geometry A < B, and A must be acute in 
order that the triangle -may be possible. If A is acute, it 
is evident from Fig. 61, 
whexeZBAC = A,AC = b, 
CB = CB' = a, that the two 
triangles ACB and ACB' 
will satisfy the given condi- 
tions, provided a is greater 
than the perpendicular CP \ 
that is, provided a is greater 
than b sin A (Sect. XI, p. 20). The angles ABC and AB'C are 
supplementary (since Z.ABC = Z.BB'C); they are, in fact, 
the supplementary angles obtained from the formula 




Fig. 61 



sin B = 



b sin A 



If, however, a = b sin A = CP (Fig. 61), then sin B = l, 
B = 90°, and the triangle required is a right triangle. 

If a < b sin A, that is, < CP, then sin B > 1, and the tri- 
angle is impossible. 



THE OBLIQUE TRIANGLE 73 

These results, for convenience, may be thus stated : 

Two solutions ; if A is acute and the value of a lies between 
b and b sin A. 

No solution ; if A is acute and a < b sin A ; 
or if A is obtuse and a < b, or a = b. 

One solution ; in all other cases. 

The number of solutions can often be determined by inspec- 
tion. In case of doubt, find the value of b sin A . 

Or we may proceed to compute log sin B. If log sin B = 0, 
the triangle required is a right triangle. If log sin B > 0, the 
triangle is impossible. If log sin B < 0, there is one solution 
when a > b ; there are two solutions when a < b. 

When there are two solutions, let B', C, c', denote the 
unknown parts of the second triangle ; then, 

B' = 180° - B, C[= 180° - (.4 + B') = B - A, 

a sin C 
c' = ■ • 

sin A 

Example 1. Given a = 16, 6 = 20, .4=106°; find the 
remaining parts. 

In this case a < b and A > 90° ; therefore, the triangle is impossible. 

Example 2. Given a = 36, b = 80, .4 = 30° ; find the 
remaining parts. 

Here we have b sin A = 80 x 4- = 40 ; so that a < b sin A and the 
triangle is impossible. 

Example 3. Given a = 72,630, b = 117,480, A =80° 0' 50" ; 
find B, C, c. 



a = 72,630 
b = 117,480 
A = 80° r 50' 



cologa = 5.13888 

log 6 = 5.06997 

log sin A = 9.99337 

loa: sin 5 = 0.20222 



Here log sin B > 0. 
.-. no solution. 



74 



PLANE TRIGONOMETRY 



Example 4. 
find B, C, c. 

a =13.2 
6 = 15.7 
A = 57° 13' 15' 



Here log sin B = 0. 
.-.a ri^M triangle. 



Example 5. 
find B, C, c. 



Given a = 13.2, 5 = 15.7, A = 57° 13' 15"; 



cologa = 8.87943 

log 6 = 1.19590 

log sin A = 9.92467 

log sin £ = 0.00000 

£=90° 

.-. C = 32° 46' 45' 



c = b cos J. 

log 6 = 1.19590 

log cos A = 9.73352 

log c = 0.92942 

c = 8.5 



Given a = 767, b = 242, A = 36° 53' 2' 



a 


= 767 


b 


= 242 


A 


= 36° 53' 2" 


Here a > 6, 


and lo 


g sin B < 0. 


.-. one 


solution. 



colog a = 7.11520 

log 6 = 2.38382 

log sin ^i = 9.77830 

log sin B- 9.27732 

5=10° 54' 58' 
.-. C= 132° 12' 0' 



log a = 2.88480 

log sin C = 9.86970 

colog sin A = 0.22170 

logc = 2.97620 

c = 946.68 



Example 6. Given a = 177.01, b 
find the other parts. 



216.45, .4=35° 36' 20"; 



a = 177.01 
b = 216:45 
A = 35° 36' 20' 



Here a<6, 
and log sin B < 0. 

.-. iiwo solutions. 



colog a = 7.75200 

log 6 = 2.33536 

log sin ^L = 9.76507 

log sin 5 = 9.85243 

£ = 45° 23' 28" 
or 134° 36' 32' 
.-. C = 99° 0' 12" 
or 9° 47' 8" 



log a =2.24800 

log sin (7 = 9.99462 

colog sin A = 0.23493 



logc = 2.47755 



2.24800 
9.23035 
0.23493 



1.71328 



300.29 or 51.675 



EXERCISE XVIH 

Find the number of solutions of the following : 

(i) a = 80, b = 100, A = 30°. 

(ii) a = 50, b = 100, . ^ = 30°. 

(iii) a = 40, b = 100, 4 = 30°. 

(iv) a = 13.4, & = 11.46, A = 77° 20'. 



THE OBLIQUE TRIANGLE 75 



(v) a = 70, 
(vi) a = 134.16, 
(vii) a = 200, 


b = 15, 
b = 84.54, 
b = 100, 


A = 60°. 

5 = 52° 9' 11". 

A = 30°. 


Given a = 840, 
find B = 12° 13' 34", 


ft = 485, 

C = 146° 15' 26" 


.4 = 21° 31'; 

, c = 1272.1. 


Given a = 9.399, 
find B = 57° 23' 40", 


ft = 9.197, 
C = 2° 1' 20", 


.1 =120° 35'; 
c = 0.38525. 


Given a = 91.06, 
find B = 41° 13', 


ft = 77.04, 
C = 87° 37' 54", 


yl =51° 9' 6": 
c = 116.82. 


Given a = 55.55, 
find .1=54° 31' 13", 


ft = 66.66, 

C = 47° 44' 7", 


£ = 77° 44' 40"; 
c = 50.481. 


Given a = 309, 


ft = 360, 


.4 = 21° 14' 25"; 



find B = 24° 57' 54", C = 133° 47' 41", c = 615.67, 
B' = 155° 2' 6", C = 3° 43' 29", c' = 55.41. 

7. Given a = 8.716, ft = 9.787, .1 = 38° 14' 12"; 

find 5 = 44°!' 28", C = 97° 44' 20", c = 1&954, 
5' = 135° 58' 32",C = 5° 47' 16", c' = 1.4202. 

8. Given a = 4.4, ft = 5.21, .1 = 57° 37' 17"; 

find B = 90°, C = 32° 22' 43", c = 2.7901. 

9. Given a = 34, ft = 22, B = 30° 20'; 

find .4 = 51° 18' 27", C = 98° 21' 33", c = 43.098, 
.4' = 128° 41' 33",C" = 20° 58' 27", c' = 15.593. 

10. Given ft = 19, c = 18, C = 15° 49'; 

find 5 = 16° 43' 13", .4 = 147° 27' 47", a = 35.519, 

B' = 163° 16' 47 ",'A = 0° 54' 13", a'= 1.0415. 

11, Given a = 75, b = 29, B = 16° 15' 36" ; find the differ- 
ence between the areas of the two corresponding triangles. 

12. Given in a parallelogram the side a, a diagonal d, and 
the angle A made by the two diagonals ; find the other diagonal. 
Special case : a = 35, d = 63, A = 21° 36' 30". 



76 PLANE TRIGONOMETRY 

SECTION XL 

Case III 

Given two sides a and b and the included angle C ; find the 
remaining parts A, B, and c. 

Solution I. The angles A and B may both be found by 
means of Formula [27], Sect. XXXVI, p. 67, which may be 
written 

tan b(A - B) = ^— X tan \{A + B). 

Since %(A + B) = £(180° - C), the value of % (A + B) is 
known, so that this equation enables us to find the value of 
$(A — B). We then have 

i(A +B) + %(A -B) = A 

and b(A + B) - ±(A - B) = B. 

After A and B are known, the side c may be found by the 
Law of Sines, which gives its value in two ways, as follows : 

a sin C b sin C 

c = — ; — — ? or c = — : 

sin A sin B 

Solution II. The third side c may be found directly from 
the equation (Sect. XXXV, p. 66) 



c = Va 2 + b 2 - 2 ab cos C ; 

and then, by the Law of Sines, the following equations for 
computing the values of the angles A and B are obtained : 

sin C . n , sin C 
sin A = a X > sm B = b X — 

(' c 



THE OBLIQUE TRIANGLE 



Solution III. If, in the triangle ABC (Fig. 62), BD is 
drawn perpendicular to the side AC, then 



tan A 




BD BD 


AD AC - DC 


Now 




BD = a sin C 


and 




DC = a cos C. 


. • . tan A 




a sin C 



b — a cos C 
By merely changing the letters, 

b sin C 




tan B = 



a — b cos C 



It is not necessary, however, to use both formulas. When 
one angle, as A, has been found, the other, B, may be found 
from the relation A + B + C = 180°. 

When the angles are known, the third side is found by the 
Law of Sines, as in Solution I. 

Note. When all three unknown parts are required, Solution I is the 
most convenient in practice. When only the third side, c, is desired, Solu- 
tion II may be used to advantage, provided the values of a 2 and b 2 can 
be obtained readily without the aid of logarithms. But Solutions II and 
III are not adapted to logarithmic work. 

Example 1. Given a = 748, b = 375, C = 63° 35' 30" ; find 
A, B, and c. 



a + b = 


1123 




a — b = 


373 




A + B = 


116° 


24' 30" 


i(A + B) = 


58° 


12' 15" 


±(A-B) = 


28° 


10' 54" 


A = 


86° 


23' 9' 


B = 


30° 


I' 21" 



log (a- 6) = 2.57171 

colog(a + 5) = 6.94962 

log tan $ (A + B) = 0.20766 

log tan ±(A -B) = 9.72899 

I (A - B) = 28° 10' 54" 



log & = 2.57403 

log sin C = 9.95214 

colog sin B = 0.30073 

log c = 2.82690 

c = 671.27 



Note. In the above example we use the angle B in finding the side 
c rather than the angle A, because A is near 90°, and therefore the use 
of its sine should be avoided. See Note, p. 23. 



78 PLANE TRIGONOMETRY 

Example 2. Given a = 4, c = 6, B = 60° ; find the third side k 

Here Solution II may be used to advantage. We have 

b = Va 2 + c 2 - 2 ac cos B = Vl6 + 36-24: = V28 ; 
log 28 = 1.44716, log V28 = 0.72358, V28 = 5.2915 ; 

that is, b = 5.2915. 

EXERCISE XIX 

1. Given a = 77.99, b = 83.39, C = 72° 15' ; 

find ,4=51° 15', B = 56° SO', c = 95.24. 

2. Given & = 872.5, c = 632.7, ^ = 80° ; 

find 5 = 60° 45' 2", C = 39° 14' 58", a = 984.83. 

3. Given a = 17, b = 12, C = 59°17'; 

find 4 = 77° 12' 53", B = 43° 30' 7", = 14.987. 

4. Given b = VE , c = V3, ^ = 35° 53' ; 

find j5 = 93° 28' 36", C = 50° 38' 24", a = 1.3131. 

5. Given a = 0.917, 5 = 0.312, C = 33°7'9"; 

find A = 132° 18' 27", B = 14° 34' 24", = 0.6775. 

6. Given a = 13.715, c = 11.214, £ = 15°22'36"; 

find^ = 118° 55' 49", C = 45° 41' 35", 5 = 4.1554. 

7. Given b = 3000.9, = 1587.2, .4 = 86° 4' 4" ; 

find 5 = 65° 13' 51", C = 28°42'5", a = 3297.2. 

8. Given a = 4527, 5 = 3465, C = 66° 6' 27"; 

find ,4=68° 29' 15", B = 45° 24' 18", c = 4449. 

9. Given a = 55.14, 5 = 33.09, C = 30°24'; 

find^ = 117°24'32", B = 32° 11' 28", = 31.431. 

10. Given a = 47.99, b = 33.14, C = 175°19'10"; 

find A =2° 46' 8", £ = 1°54'42", c = 81.066. 

11. If two sides of a triangle are each equal to 6, and the 
included angle is 60°, find the third side. 



THE OBLIQUE TRIANGLE 79 

12. If two sides of a triangle are each equal to 6, and the 
included angle is 120°, find the third side. 

13. Apply Solution I to the case in which a is equal to b ; 
that is, the case in which the triangle is isosceles. 

14. If two sides of a triangle are 10 and 11, and the included 
angle is 50°, find the third side. 

15. If two sides of a triangle are 43.301 and 25, and the 
included angle is 30°, find the third side. 

16. In order to find the distance between two objects, A 
and B, separated by a swamp, a station C was chosen, and the 
distances CA = 3825 yards, CB = 3475.6 yards, together with 
the angle A CB = 62° 31', were measured. Find the distance 
from A to B. 

17. Two inaccessible objects, A and B, are each viewed from 
two stations, C and D, on the same side of AB and 562 yards 
apart. The angle ACB is 62° 12', BCD 41° 8', ADB 60° 49', 
and ADC 34° 5.1' ; required the distance AB. 

18. Two trains start at the same time from the same station 
and move along straight tracks that form an angle of 30°, one 
train at the rate of 30 miles an hour, the other at the rate of 
40 miles an hour. How far apart are the trains at the end of 
half an hour ? 

19. In a parallelogram given the two diagonals 5 and 6, 
and the angle that they form 49° 18'; find the sides. 

20. In a triangle one angle is 139° 54', and the sides forming 
the angle have the ratio 5 : 9. Find the other two angles. 

21. In order to find the distance between two objects, A 
and B, separated by a pond, a station C was chosen, and the 
distances CA = 426 yards, CB = 322.4 yards, together with 
the angle ACB = 68° 42', were measured. Find the distance 
from A to B. 



80 



PLANE TRIGONOMETRY 



SECTION XLI 



Case IY 

Given the three sides a, b, c ; find the angles A, B, C. 

The angles may be found directly from the formulas estab- 
lished in Sect. XXXV, p. 66. Thus, from the formula 

a 2 = b 2 + c 2 — 2 be cos A, 

b 2 + c 2 - a 2 



cos A 



2 be 



From this equation formulas adapted to logarithmic work 
are deduced as follows : 

For the sake of brevity, let 

a + b + c = 2s; 

then b + e — a = 2 (s — a), 

D B a -b + e = 2(s-b), 

fig. 63 and a + b — e = 2{s —c). 

Then the value of 1 — cos A is 

tf + c 2 -a 2 2be-b 2 -c 2 + a 2 c 




(b-cY 



2 be 2 be 

_ (a + b — c)(a — b + c) 
" 2be 

2(s — b)(s-c) 

be 

and the value of 1 + cos A is 

£2 + c 2 _ a 2 2bc + b 2 + C 2 -C 



2 be 



1 + 



(b + ey 



2 be 



2 be 



2 be 



_ (b + c + a) (b + c — a) _ 2 s (s — a) 
~ 2 be ~ be 



THE OBLIQUE TRIANGLE 81 

But from Formulas [16] and [17], p. 58, it follows that 
1 — cos A = 2 sin 2 £ A, and 1 + cos A = 2 cos 2 £ A. 

.-, a «mH,l~ a '£ -'><'-'> , and 2cos^,l=^^>, 
be be 



whence sin £ A = vi ii 1 , 

* be 



cosjA = J s f s -- a ), 
v be 



[28] 
[29] 



and by [2], tan | A = J (8-b)(8-c) _ ^ 

S (S — SI ) 

By merely changing the letters, 




sin # c = ^« -«)(*-*) 



cos+c = \fc^ 

2 * aft 



s(s-b) X s(s-c) 



tan i - 



There is then a choice of three different formulas for finding 
the value of each angle. If half the angle is very near 0°, 
the formula for the cosine will not give a very accurate result, 
because the cosines of angles near 0° differ little in value ; and 
the same holds true of the formula for the sine when half the 
angle is very near 90°. Hence, in the first case the formula 
for the sine, in the second that for the cosine, should be used. 

But, in general, the formulas for the tangent are to be 
preferred. 

It is not necessary to compute by the formulas more than 
two angles ; for the third may then be found from the equation 

A + B + C = 180°. 



82 



PLANE TRIGONOMETRY 



There is this advantage, however, in computing all three 
angles by the formulas, that we may then use the sum of the 
angles as a test of the accuracy of the results. 

In case it is desired to compute all the angles, the formulas 
for the tangent may be put in a more convenient form. 

The value of tan £ A may be written 



or 





xF 


-a)(s 


-*)(«- 


») 




V 


s(s 


-a? 






1 J(» 


- a)(s 


-b)(s- 


«) 


Hence, 


s — a ^ 

if we put 




s 






vp 1 


)(s_b)(s-c) 


= r, 



we have 

Likewise, tan \ B = 



tan^ A 



— a 



tan \ C = 



[31] 
[32] 



Example 1. Given a = 3.41, b = 2.60, c = 1.58; find the 
angles. 

Using Formula [30] and the corresponding formula for tan £ B, we 
may arrange the work as follows : 



a = 3.41 
6 = 2.60 
c = 1.58 
2s = 7.59 
8 = 3.795 
s _ a = 0.385 
s - b = 1.195 
s - c = 2.215 



colog s = 9.42079 

colog (s — a) = 0.41454 

log (s - b) = 0.07737 

log (s-c) = 0.34537 

2 )0.25807 

log tan £ J. =0.12903 

i^L= 53° 23' 20' 
A = 106° 46' 40' 



colog s= 9.42079- 10 
log(s-a) = 9.58546- 10 
colog (s -b)= 9.92263 - 10 
log (s - c) = 0.34537 

2 )19.27425- 20 
logtan£i?= 9.63713-10 
$B = 2S° 26' 37" 
B = 46° 53' 14" 



.-. A + B = 153° 39' 54", and C = 26° 20' 6". 



THE OBLIQUE TRIANGLE 



83 



Example 2. Solve Example 1 by finding all three angles 
by the use of Formulas [31] and [32], 

Here the "work may be compactly arranged as follows, if we find 
log tan £.4, etc., by subtracting log(s — a), etc., from logr instead of 
adding the cologarithm : 



a = 3.41 

6 = 2.60 

c = 1.58 

2s = 7.59 



s = 3.795 
s-a = 0.385 
s- 6 = 1.195 
s- c = 2.215 

2 s = 7.590 (check). 



log(s -a) = 9.58546 

log (s -b) = 0.07737 

log(s - c) = 0.34537 

colog s = 9.42079 

logr 2 



9.42899 
logr =9.71450 



log tan | A = 10.12903 

log tan \ B = 9.63713 

log tan i C = 9.36912 

\A = 

$B = 

}C = 

^L = 

B = 

C = 



53° 


23' 20" 


23° 


26' 37" 


13° 


10' 3" 


106° 


46' 40" 


46° 


53' 14" 


26° 


20' 6" 



Check, A + -B + C = 180° 0' 0" 

Note. Even if no mistakes are made in the work, the sum of the 
three angles found as above may differ very slightly from 180° in conse- 
quence of the fact that logarithmic computation is at best only a method 
of close approximation. When a difference of this kind exists, it should 
be divided among the angles according to the probable amount of error 
for each angle. 

EXERCISE XX 

Solve the following triangles, taking the three sides as the 
given parts : 



1 


a 


b 


c 


A 


D 


1 
C 


51 


65 


20 


38° 52' 48" 


126° 52' 12" 


14° 15 r 


2 


78 


101 


29 


32° 10' 55" 


136° 23' 50" 


11° 25' 15" 


3 


111 


145 


40 


27° 20' 32" 


143° 7' 48" 


9° 31' 40" 


4 


21 


26 


31 


42° 6' 13" 


56° 6' 36" 


81° 47' 11" 


5 


19 


34 


49 


16° 25' 36" 


30° 24' 


133° 10' 24" 


6 


43 


50 


57 


46° 49' 35" 


57° 59' 44" 


75° 10' 41" 


7 


37 


58 


79 


26° 0' 29" 


43° 25' 20" 


110° 34' 11" 


8 


73 


82 


91 


49° 34' 58" 


58° 46' 58" 


71° 38' 4" 


9 


14.493 


55.4363 


66.9129 


8° 20' 1" 


33° 40' 5" 


137° 59' 54" 


10 


V5 


V6~ 


V7 


51° 53' 12" 


59° 31' 48" 


68° 35' 



84 PLANE TRIGONOMETRY 

\ll. Given a = 6, b = 8, c = 10; find the angles. 

12. Given a = 6, b = 6, c = 10 ; find the angles. 

13. Given a = 6, b = 6, c = 6 ; find the angles. 

14. Given a = 6, 6 = 9, c = 12 ; find the angles. 

15. Given a = 2, b = V6, c = V3 — 1 ; find the angles. 

16. Given a = 2, b = V6, c = V3 + 1 ; find the angles. 



\ 



17. The distances between three cities, A, B, and C, are as 
follows : AB = 165 miles, AC = 12 miles, and BC = 185 miles. 
B is due east from A. In what direction is C from A ? What 
two answers are admissible ? 

. 18. Under what visual angle is an object 7 feet long seen 
Hby an observer whose eye is 5 feet from one end of the object 
and 8 feet from the other end ? 

19. When Formula [28] is used for finding the value of an 
angle, why does the ambiguity that occurs in Case II not exist ? 

20. If the sides of a triangle are 3, 4, and 6, find the sine of 
the largest angle. 

21. Of three towns, A, B, and C, A is 200 miles from B and 
184 miles from C, B is 150 miles due north from C. How far 
is A north of C ? 

22. The sides of a triangle are 78.9, 65.4, 97.3, respectively. 
Find the largest angle. 

23. The sides of a triangle are 487.25, 512.33, 544.37, 
xrespectively. Find the smallest angle. 

T^ -r.. -. i , o • n l l V3 + 1 

24. Find the angles of a triangle whose sides are y=- > 

V3-1 V3 + . n 2V ^ 

y- > np respectively. 

25. The sides of a triangle are 14.6 inches, 16.7 inches, and 
18.8 inches, respectively. Find the length of the perpendic- 
ular from the vertex of the largest angle upon the opposite side. 



THE OBLIQUE TRIANGLE 



85 



SECTION XLII 
AREA OF A TRIANGLE 

Case I 
When two sides and the included angle are given. 





In the triangle ABC (Fig. 64 or 65), 
F = %cx CD. 
Now, CD = a sin B. 

Therefore, F = -J- ac sin B. 

Also, F = \ab sin C, and F = %bc sin A. 



[33] 



Case II 
When a side and the two adjacent angles are given. 

sin A : sin C = a : c. (Sect. XXXIV, p. 65.) 



Therefore, 



c = 



a sin C 
sin A 



Putting this value of c in Formula [33], 

a 2 sin B sin C 



F = 



2 sin A 



But sin (i? + C) = sin (180° - A) = sin .4. (Sect. XXV, p. 48.) 



Hence, 



F = 



a 2 sin B sin C 

2 sin (B + C) 



[34] 



86 



PLANE TRIGONOMETRY 



Case III 

Wh en the three sides of a triangle are given. 
By Formula [12], p. 58, 

sin B = 2 sin -J -B X cos \ B. 
Now, by Formula [28], p. 81, 

• i t* - K s — a)(s — c) 
* ac 

and by Formula [29], p. 81, 

cos \ B = yj s ^ ~ ^ • 

By substituting these values of sin \ B and cos \ B in the 
above equation, we have 

2_ 
ac 



sin B = — Vs (s — a)(s — ft) (s — c). 
etc 

By putting this value of sin i? in [33], we have 



F = Vs (s — a) (s — b) (s — c). 



[35] 



Case IV 

When the three sides and the radius of the circumscribed 
circle or the radius of the inscribed circle are given. 

A If i? denotes the radius of the 

circumscribed circle, we have, from 
Sect. XXXIV, p. 65, 
b 




smB = 



2R 



By putting this value of sin B 
in [33], we have 



* = £• [36] 



Fig. 66 



THE OBLIQUE TRIANGLE 87 

If r denotes the radius of the inscribed circle, divide the 
triangle into three triangles by lines from the centre of this 
circle to the vertices ; then the altitude of each of the three 
triangles is equal to r. Therefore, 

F = £r(a + b + c) = rs. [37] 

By putting in this formula the value of F given in [35], 



# 



a) (s — b) (s — c) 



whence r, in [31], Sect. XLI, p. 82, is equal to the radius of 
the inscribed circle. 



EXERCISE XXI 




Find the area : 






1. Given a = 4474.5, 


b = 2164.5, 


C = 116° 30' 20". 


2. Given b = 21.66, 


c = 36.94, 


A = 66° 4' 19". 


3. Given a = 510, 


c = 173, 


B = 162° 30' 28". 


4. Given a = 408, 


5 = 41, 


c = 401. 


5. Given a = 40, 


b = 13, 


c = 37. 


6. Given a = 624, 


b = 205, 


c = 445. 


7. Given b = 149, 


A = 70° 42' 30", 


5 = 39° 18' 28". 


8. Given a = 215.9, 


c = 307.7, 


4 = 25° 9' 31". 


9. Given b = 8, 


c = 5, 


A = 60°. 


10. Given a = 7, 


c = 3, 


A = 60°. 



11. Given a = 60, 5 = 40° 35' 12", area = 12 ; find the 
radius of the inscribed circle. 

12. Obtain a formula for the area of a parallelogram in 
terms of two adjacent sides and the included angle. 

13. Obtain a formula for the area of an isosceles trapezoid 
in terms of the two parallel sides and an acute angle. 



88 PLANE TRIGONOMETRY 

14. Two sides and included angle of a triangle are 2416, 
1712, and 30° ; and two sides and included angle of another 
triangle are 1948, 2848, and 150°. Find the sum of their areas. 

15. The base of an isosceles triangle is 20, and its area is 
100 -=- V3 ; find its angles. 

16. Show that the area of a quadrilateral is equal to one- 
half the product of its diagonals into the sine of their included 
angle. 

EXERCISE XXII 

1. From a ship sailing down the English Channel the Eddy- 
stone was observed to bear N. 33° 45' W., and after the ship 
had sailed 18 miles S. 67° 30' W. it bore N. 11° 15' E. Eind 
its distance from each position of the ship. 

2. Two objects, A and B, were observed from a ship to be 
at the same instant in a line bearing N. 15° E. The ship then 
sailed northwest 5 miles, when it was found that A bore due 
east and B bore northeast. Eind the distance from A to B. 

3. A castle and a monument stand on the same horizontal 
plane. The angles of depression of the top and the bottom of 
the monument viewed from the top of the castle are 40° and 
80° ; the height of the castle is 140 feet. Eind the height of 
the monument. 

4. If the sun's altitude is 60°, what angle must a stick make 
with the horizon in order that its shadow in a horizontal plane 
may be the longest possible ? 

5. If the sun's altitude is 30°, find the length of the longest 
shadow cast on a horizontal plane by a stick 10 feet in length. 

6. In a circle with the radius 3 find the area of the part 
comprised between parallel chords whose lengths are 4 and 5. 
(Two solutions.) 



CHAPTER V 

MISCELLANEOUS EXAMPLES 

Problems in Plane Trigonometry 




/. / : i 



ft --'J'jJ 




Fig. 67 

If two objects are not in the same horizontal plane with 
each other or with the point of observation, we may suppose 
vertical lines to be passed through the two objects and to 
meet the horizontal plane of the point of observation in two 
points. The angular distance of these two points is the bear- 
ing of either of the objects from the other. Thus, the angle 
COD' (Fig. 67) is the bearing of C from D. 

Note. "Problems in Plane Trigonometry " are selected from those 
published by Mr. Charles W. Seaver, Cambridge, Mass. The full set can 
be obtained from him in pamphlet form. 

89 



90 PLANE TRIGONOMETRY 

EXERCISE XXIII 
RIGHT TRIANGLES 

1. The angle of elevation of a tower is 48° 19' 14", and 
the distance of the base from the point of observation is 
95 feet. Find the height of the tower and the distance of 
its top from the point of observation. 

2. From a mountain 1000 feet high, the angle of depression 
of a ship is 77° 35' 11". Find the distance of the ship from 
the summit of the mountain. 

3. A flagstaff 90 feet high, on a horizontal plane, casts a 
shadow of 117 feet. Find the altitude of the sun. 

4. When the moon is setting at any place, the angle at the 
moon subtended by the earth's radius passing through that 
place is 57' 3". If the earth's radius is 3956.2 miles, what is 
the moon's distance from the earth's centre ? 

5. The angle at the earth's centre subtended by the sun's 
radius is 16' 2" and the sun's distance is 92,400,000 miles. 
Find the sun's diameter in miles. 

6. The latitude of Cambridge, Mass., is 42° 22' 49". What 
is the length of the radius of that parallel of latitude ? 

7. At what latitude is the circumference of the parallel of 
latitude half of that of the equator ? 

8. In a circle with a radius of 6.7 is inscribed a regular 
polygon of thirteen sides. Find the length of one of its sides. 

9. A regular heptagon, one side of which is 5.73, is 
inscribed in a circle. Find the radius of the circle. 

10. A tower 93.97 feet high is situated on the bank of a 
river. The angle of depression of an object on the opposite 
bank is 25° 12' 54". Find the breadth of the river. 



MISCELLANEOUS EXAMPLES 91 

11. From a tower 58 feet high the angles of depression of 
two objects situated in the same horizontal line with the 
base of the tower, and on the same side, are 30° 13' 18" and 
45° 46' 14". Find the distance between these two objects. 

12. Standing directly in front of one corner of a flat-roofed 
house, which is 150 feet in length, I observe that the hori- 
zontal angle which the length subtends has for its cosine Vi, 

and that the vertical angle subtended by its height has for its 

3 
sine — -p=' What is the height of the house ? 
V34 

13. A regular pyramid, with a square base, has a lateral 
edge 150 feet long, and a side of its base is 200 feet. Find 
the inclination of the face of the pyramid to the base. 

14. From one edge of a ditch 36 feet wide, the angle of 
elevation of a wall on the opposite edge is 62° 39' 10". Find 
the length of a ladder that will just reach from the point 
of observation to the top of the wall. 

^t&r The top of a flagstaff has been partly broken off and 
touches the ground at a distance of 15 feet from the foot of 
the staff. If the length of the broken part is 39 feet, find the 
length of the whole staff. 

16. From a balloon, which is directly above one town, is 
observed the angle of depression of another town, 10° 14' 9". 
The towns being 8 miles apart, find the height of the balloon. 

17. From the top of a mountain 3 miles high the angle of 
depression of the most distant object which is visible on the 
earth's surface is found to be 2° 13' 50". Find the diameter 
of the earth. 

^Sr^ATTadder 40 feet long reaches a window 33 feet high, 
on one side of a street. Being turned over upon its foot, it 
reaches another window 21 feet high, on the opposite side of 
the street. Find the width of the street. 



92 PLANE TRIGONOMETRY 

19. The height of a house subtends a right angle at a 
window on the other side of the street; and the angle of 
elevation of the top of the house, from the same point, is 60°. 
The street is 30 feet wide. How high is the house ? 

20. A lighthouse 54 feet high is situated on a rock. The 
angle of elevation of the top of the lighthouse, as observed 
from a ship, is 4° 52', and the angle of elevation of the top of 
the rock is 4° 2'. Find the height of the rock and its distance 
from the ship. 

21. A man in a balloon observes the angle of depression of 
an object on the ground, bearing south, to be 35° 30'; the 
balloon drifts 2\ miles east at the same height, when the angle 
of depression of the same object is 23° 14'. Find the height 
of the balloon. 

22. A man standing south of a tower, on the same horizon- 
tal plane, observes its angle of elevation to be 54° 16' ; he goes 
east 100 yards, and then finds its angle of elevation is 50° 8'. 
Find the height of the tower. 

23. The angle of elevation of a tower at a place A, south of 
it, is 30° ; and at a place B, west of A, and at a distance of a from 
it, the angle of elevation is 18°. Show that the height of the 

tower is —-==== ; the tangent of 18° being 



V2 4- 2 V5 V10 + 2V5 

24. A pole is fixed on the top of a mound, and the angles 
of elevation of the top and the bottom of the pole are 60° and 
30°, respectively. Prove that the length of the pole is twice 
the height of the mound. 

25. At a distance a from the foot of a tower, the angle 
of elevation A of the top of the tower is the complement of 
the angle of elevation of a flagstaff on top of it. Show that 
the length of the staff is 2 a cot 2 A. 



MISCELLANEOUS EXAMPLES 93 

26. A line of true level is a line every point of which is 
equally distant from the centre of the earth. A line drawn 
tangent to a line of true level at any point is a line of 
apparent level. If at any point both these lines are drawn, 
and extended one mile, find the distance they are then apart. 

27. In Problem 1, page 90, determine the effect upon the 
computed height of the tower of an error in either the angle 
of elevation or the measured distance. 



OBLIQUE TRIANGLES 

28. To determine the height of an inaccessible object 
situated on a horizontal plane, by observing its angles of 
elevation at two points in the same line with its base, and 
measuring the distance between these two points. 

29. The angle of elevation of an inaccessible tower situated 
on a horizontal plane is 63° 26'; at a point 500 feet farther 
from the base of the tower the angle of elevation of its top is 
32° 14'. Find the height of the tower. 

-30. A tower is situated on the bank of a river. From the 
opposite bank the angle of elevation of the tower is 60° 13', 
and from a point 40 feet more distant the angle of elevation 
is 50° 19'. Find the breadth of the river. 

31. A ship sailing north sees two lighthouses 8 miles apart, 
in a line due west ; after an hour's sailing, one lighthouse 
bears S.W., and the other S.S.W. Find the ship's rate. 

32. To determine the height of an accessible object situated 
on an inclined plane. 

33. At the distance of 40 feet from the foot of a tower on 
an inclined plane, the tower subtends an angle of 41° 19'; at 
a point 60 feet farther away, the angle subtended by the tower 
is 23° 45'. Find the height of the tower. 



94 PLANE TRIGONOMETRY 

34. A tower makes an angle of 113° 12' with the inclined 
plane on which it stands ; and at a distance of 89 feet from its 
base, measured down the plane, the angle subtended by the 
tower is 23° 27'. Find the height of the tower. 

35. From the top of a house 42 feet high the angle of 
elevation of the top of a pole is 14° 13'; at the bottom of 
the house it is 23° 19'. Find the height of the pole. 

36. The sides of a triangle are 17, 21, 28. Prove that the 
length of a line bisecting the greatest side and drawn from 
the opposite angle is 13. 

37. A privateer, 10 miles S.W. of a harbor, sees a ship sail 
from it in a direction S. 80° E., at a rate of 9 miles an hour. 
In what direction, and at what rate, must the privateer sail 
in order to come up with the ship in 1J hours ? 

38. A person goes 70 yards up a slope of 1 in 3|- from the 
edge of a river and observes the angle of depression of an 
object on the opposite bank to be 2^°. Find the breadth of 
the river. 

39. The length of a lake subtends, at a certain point, an 
angle of 46° 24', and the distances from this point to the two 
extremities of the lake are 346 and 290 feet. Find the length 
of the lake. 

40. Two ships are a mile apart. The angular distance of 
the first ship from a fort on shore, as observed from the second 
ship, is 35° 14' 10"; the angular distance of the second ship 
from the fort, observed from the first ship, is 42° 11' 53". 
Find the distance in feet from each ship to the fort. 

41. Along the bank of a river is drawn a base line of 500 
feet. The angular distance of one end of this line from an 
object on the opposite side of the river, as observed from the 
other end of the line,, is 53° ; that of the second extremity 



MISCELLANEOUS EXAMPLES 95 

from the same object, observed at the first, is 79° 12'. Find 
the breadth of the river. 

Jj&f^K vertical tower stands on a declivity inclined 15° to 
the horizon. A man ascends the declivity 80 feet from the 
base of the tower, and finds the angle then subtended by the 
tower to be 30°. Find the height of the tower. 

43. The angle subtended by a tower on an inclined plane is, 
at a certain point, 42° 17'; 325 feet farther down it is 21° 47'. 
The inclination of the plane is 8° 53'. Find the height of the 
tower. 

44. A cape bears north by east, as seen from a ship. The 
ship sails northwest 30 miles, and then the cape bears east. 
How far is it from the second point of observation? 

45. Two observers, stationed on ojiposite sides of a cloud, 
observe its angles of elevation to be 44° 50' and 36° 4'. Their 
distance from each other is 700 feet. What is the height of 
the cloud ? 

46. From a point B at the foot of a mountain, the angle of 
elevation of the top A is 60°. After ascending the mountain 
one mile, at an inclination of 30° to the horizon, and reaching 
a point C, the angle A CB is found to be 135°. Find the height 
of the mountain in feet. 

47. From a ship two rocks are seen in the same right line 
with the ship, bearing N. 15° E. After the ship has sailed 
northwest 5 miles, the first rock bears east, and the second 
northeast. Find the distance between the rocks. 

/4d< < ^rom a window on a level with the bottom of a steeple 
the angle of elevation of the steeple is 40°, and from a second 
window 18 feet higher the angle of elevation is 37° 30'. Find 
the height of the steeple. 




96 PLANE TRIGONOMETRY 

49. To determine the distance between two inaccessible 
objects by observing angles at the extremities of a line of 
known length. 

50. Wishing to determine the distance between a church A 
and a tower B, on the opposite side of a river, I measure a 
line CD along the river (C being nearly opposite A), and 
observe the angles A CB, 58° 20' ; A CD, 95° 20' ; ADB, 53° 30'; 
BDC, 98° 45'. CD is 600 feet. What is the distance required? 

51. Wishing to find the height of a summit A, I measure a 
horizontal base line CD, 440 yards. At C, the angle of eleva- 
tion of A is 37° 18', and the horizontal angle between D and 
the summit is 76° 18' ; at D, the horizontal angle between C 
and the summit is 67° 14'. Find the height. 

52. A balloon is observed from two stations 3000 feet apart. 
At the first station the horizontal angle of the balloon and the 
other station is 75° 25', and the angle of elevation of the balloon 
is 18°. The horizontal angle of the first station and the bal- 
loon, measured at the second station, is 64° 30'. Find the 
height of the balloon. 

53. Two forces, one of 410 pounds, and the other of 320 
pounds, make an angle of 51° 37'. Find the intensity and the 
direction of their resultant. 

54. An unknown force, combined with one of 128 pounds, 
produces a resultant of 200 pounds, and this resultant makes 
an angle of 18° 24' with the known force. Find the intensity 
and direction of the unknown force. 

55. At two stations, the height of a kite subtends the same 
angle A. The angle which the line joining one station and 
the kite subtends at the other station is B ; and the distance 
between the two stations is a. Show that the height of the 
kite is i a sin A sec B. 



4 



MISCELLANEOUS EXAMPLES 97 



56. Two towers on a horizontal plane are 120 feet apart. A 
person standing successively at their bases observes that the 
angle of elevation of one is double that of the other; but, 
when he is half-way between them, the angles of elevation are 
complementary. Prove that the heights of the towers are 90 
and 40 feet. 

57. To find the distance of an inaccessible point C from 
either of two points A and B, having no instruments to meas- 
ure angles. Prolong CA to a, and CB to b, and join AB, Ab, 
and Ba. Measure AB, 500 ; a A, 100 ; aB, 560 ; bB, 100 ; and 
Ab, 550. Compute the distances AC and BC. 

58. Two inaccessible points A and B are visible from D, 
but no other point can be found whence both are visible. 
Take some point C, whence A and D can be seen, and meas- 
ure CD, 200 feet; ADC, $9°; A CD, 50° 30'. Then take some 
point E, whence D and B are visible, and measure DE, 200 
feet; BDE, 54° 30' ; BED, 88° 30'. At D measure ADB, 72° 30'. 
Compute the distance AB. 

59. To compute the horizontal distance between two inac- 
cessible points A and B, when no point can be found whence 
both can be seen. Take two points C and D, distant 200 yards, 
so that A can be seen from C, and B from D. From C meas- 
ure CF, 200 yards to F, whence A can be seen ; and from D 
measure DE, 200 yards to E, whence B can be seen. Measure 
AFC, 83°; ACD, 53° 30'; ACF, 54° 31'; BDE, 54° 30'; BDC, 
156° 25'; DEB, 88° 30'. 

60. A column in the north temperate zone is east-southeast 
of an observer, and at noon the extremity of its shadow is 
northeast of him. The shadow is 80 feet in length, and the 
elevation of the column, at the observer's station, is 45°. 
Find the height of the column. 



98 PLANE TRIGONOMETRY 

61. From the top of a hill the angles of depression of two 
objects situated in the horizontal plane of the base of the hill 
are 45° and 30°; and the horizontal angle between the two 
objects is 30°. Show that the height of the hill is equal to 
the distance between the objects. 

62. Wishing to know the breadth of a river from A to B, I 
take AC, 100 yards in the prolongation of BA, and then take 
CD, 200 yards at right angles to AC. The angle BDA is 
37° 18' 30". Find AB. 

63. The sum of the sides of a triangle is 100. The angle at 
A is double that at B, and the angle at B is double that at C. 
Determine the sides. 

64. If sin 2 A + 5 cos 2 A = 3, find A. 

65. If sin 2 J. = m cos A — n, find cos A. 

66. Given sin A—m sin B, and tan A — n tan B ; find sin A 
and cos B. 

67. If ton* A + 4 sin 2 4 = 6, find A. 
6-8. If sin A = sin 2 A, find A. 

69. If tan 2 A = 3 tan A, find A. 

70. Prove that tan 50° + cot 50° = 2 sec 10°. 

71. Given a regular polygon of n sides, and calling one of 
them a, find expressions for the radii of the inscribed and the 
circumscribed circles in terms of n and a. 

If P, H, D are the sides of a regular inscribed pentagon, 
hexagon, and decagon, prove P 2 = H 2 + D*. 

AREAS 

72. Obtain the formula for the area of a triangle, given two 
sides b, c, and the included angle A. 

73. Obtain the formula for the area of a triangle, given two 
angles A and B, and included side c. 



MISCELLANEOUS EXAMPLES 99 

74. Obtain the formula for the area of a triangle, given the 
three sides. 

75. If a is the side of an equilateral triangle, show that 

. a 2 V3 

its area is — ; 

4 

76. Two consecutive sides of a rectangle are 52.25 chains 
and 38.24 chains. Find the area. 

77. Two sides of a parallelogram are 59.8 chains and 37.05 
chains, and the included angle is 72° 10'. Find the area. 

78. Two sides of a parallelogram are 15.36 chains and 11.46 
chains, and the included angle is 47° 30'. Find the area. 

79. Two sides of a triangle are 12.38 chains and 6.78 chains, 
and the included angle is 46° 24'. Find the area. 

80. Two sides of a triangle are 18.37 chains and 13.44 
chains, and they form a right angle. Find the area. 

81. Two angles of a triangle are 76° 54' and 57° 33' 12", 
and the included side is 9 chains. Find the area. 

82. Two sides of a triangle are 19.74 chains and 17.34 
chains. The first bears K 82° 30' W. ; the second S. 24° 15' E. 
Find the area. 

83. The three sides of a triangle are 49 chains, 50.25 chains, 
and 25.69 chains. Find the area. 

84. The three sides of a triangle are 10.64 chains, 12.28 
chains, and 9 chains. Find the area. 

85. The sides of a triangular field, of which the area is 14 
acres, are in the ratio of 3, 5, 7. Find the sides. 

86. In the quadrilateral ABCD we have AB, 17.22 chains; 
AD, 7.45 chains ; CD, 14.10 chains ; BC, 5.25 chains ; and the 
diagonal AC, 15.04 chains. Find the area. 

L.ofC. 



100 PLANE TRIGONOMETRY 

87. The diagonals of a quadrilateral are a and b, and they 
intersect at an angle D. Show that the area of the quadri- 
lateral is \ ab sin D. 

88. The diagonals of a quadrilateral are 34 and 56, inter- 
secting at an angle of 67°. Find the area. 

89. The diagonals of a quadrilateral are 75 and 49, inter- 
secting at an angle of 42°. Find the area. 

90. Show that the area of a regular polygon of n sides, of 

. no? t 180° 

which one is a, is —r- cot 

4 n 

91. One side of a regular pentagon is 25. Find the area. 

92. One side of a regular hexagon is 32. Find the area. 

93. One side of a regular decagon is 46. Find the area. 

94. Find the area of a circle whose circumference is 74 feet. 

95. Find the area of a circle whose radius is 125 feet. 

96. In a circle with a diameter of 125 feet find the area of 
a sector with an arc of 22°. 

97. In a circle with a radius of 44 feet find the area of a 
sector with an arc of 25°. 

98. In a circle with a diameter of 50 feet find the area of 
a segment with an arc of 280°. 

99. Find the area of a segment (less than a semicircle) of 
which the chord is 20, and the distance of the chord from the 
middle point of the smaller arc is 2. 

100. If r is the radius of a circle, the area of a regular 

180° 
circumscribed polygon of n sides is nr 1 tan 

The area of a regular inscribed polygon is » r 2 sin 

— TV 



MISCELLANEOUS EXAMPLES 101 

101. If a is a side of a regular polygon of n sides, the area 

of the inscribed circle is —j- cot 2 

The area of the circumscribed circle is — — esc 2 



102. The area of a regular polygon inscribed in a circle is 
to that of the circumscribed regular polygon of the same 
number of sides as 3 to 4. Find the number of sides. 

103. The area of a regular polygon inscribed in a circle is 
the geometric mean between the areas of an inscribed and a 
circumscribed regular polygon of half the number of sides. 

104. The area of a circumscribed regular polygon is the 
harmonic mean between the areas of an inscribed regular 
polygon of the same number of sides and of a circumscribed 
regular polygon of half that number. 

105. The perimeter of a circumscribed regular triangle is 
double that of the inscribed regular triangle. 

106. The square described about a circle is four-thirds the 
inscribed regular dodecagon. 

107. Two sides of a triangle are 3 and 12, and the included 
angle is 30°. Find the hypotenuse of an isosceles right tri- 
angle of equal area. 

PLANE SAILING 

Plane Sailing is that branch of Navigation in which the 
surface of the earth is considered a plane. The problems 
which arise are therefore solved by the methods of Plane 
Trigonometry. 

The difference of latitude of two places is the arc of a 
meridian comprehended between the parallels of latitude 
passing through those places. 



102 PLANE TRIGONOMETRY 

The departure between two meridians is the arc of a parallel 
of latitude comprehended between those meridians. It dimin- 
ishes as the distance from the equator increases. 

When a ship sails in such a manner as to cross successive 
meridians at the same angle, it is said to sail on a rhumb-line. 
This angle is called the course, and the distance between two 
places is measured on a rhumb-line. 

If we consider the distance, departure, and difference of 
latitude of two places to be straight lines, lying in one plane, 
they form a right triangle, called the triangle of plane sailing. 
If ABC is a plane triangle, right-angled at B, and BC repre- 
sents the difference of latitude of B and C, ACB will be the 
course from C to A, CA the distance, and D the departure, 
measured from B, between the meridian of A and that of B. 

108. Taking the earth's equatorial diameter to be 7925.6 
miles, find the length in feet of the arc of one minute of a 
great circle.* 

109. A ship sails from latitude 43° 45' S., on a course 
N. by E. 2345 miles. Find the latitude reached, and the 
departure made. 

110. A ship sails from latitude 1° 45' N., on a course S.E. 
by E., and reaches latitude 2° 31' S. Find the distance, and 
the departure. 

111. A ship sails from latitude 13° 17' S., on a course N.E. 
by E. f E., until the departure is 207 miles. Find the distance, 
and the latitude reached. 

112. A ship sails on a course between S. and E. 244 miles, 
leaving latitude 2° 52' S., and reaching latitude 5° 8' S. Find 
the course, and the departure. 

* The length of the arc of one minute of a great circle of the earth is 
called a geographical mile or a knot. In the following problems, this is 
the distance meant by the term "mile," unless otherwise stated. 



MISCELLANEOUS EXAMPLES 



103 



113. A ship sails from latitude 32° 18' X., on a course 
between N. and W., a distance of 344 miles, and a departure 
of 103 miles. Find the course, and the latitude reached. 

114. A ship sails on a course between S. and E., making a 
difference of latitude 136 miles, and a departure 203 miles. 
Find the distance, and the course. 

115. A ship sails due north 15 statute miles an hour, for 
one day. What is the distance, in a straight line, from the 
point left to the point reached ? (Take earth's radius, 3962.8 
statute miles.) 

PARALLEL AND MIDDLE LATITUDE SAILING 

The difference of longitude of two places is the angle at the 
pole made by the meridians of these two places ; or, it is the 
arc of the equator comprehended between these two meridians. 





Fig. G6 

In Parallel Sailing a vessel is supposed to sail due east or 
due west. The distance sailed is the departure made ; and 
the difference of longitude is found as follows : 

116. Given the departure between any two meridians at 
any latitude ; find the difference of longitude of any point on 
one meridian from any point on the other. 



104 



PLANE TRIGONOMETRY 




In rt. A ODA, Z A OD = 90 
DA 



lat. 



Hence, 



OA 



= sin (90° — lat.) = cos lat. 



The A DAB and OEQ are similar. 



Therefore, 



DA AB 



OE 
Hence, cos lat. 



Therefore, EQ = 
That is 



EQ 
AB 

eq' 

AB 



DA 

or 

OA 



AB 
EQ 



= AB xsecl&t. 



cos lat. 
Diff . long. = depart, x sec lat. 

117. A ship in latitude 42° 16' K, longitude 72° 16' W., 
sails due east a distance of 149 miles. What is the position 
of the point reached ? 

118. A ship in latitude 44° 49' S., longitude 119° 42' E., 
sails due west until it reaches longitude 117° 16' E. Find 
the distance made. 

In Middle Latitude Sailing the departure between two places 
is measured on that parallel of latitude which lies midway 
between the parallels of the two places. Except in very high 
latitudes or excessive runs, this assumption produces no great 
error. Hence, in middle latitude sailing, 

Diff. long. = depart, x sec mid. lat. 




Fig. 71 




MISCELLANEOUS EXAMPLES 105 

119. A ship leaves latitude 31° 14' N., longitude 42° 19' W., 
and sails E.N.E. 325 miles. Find the position reached. 

120. Find the bearing and distance of Cape Cod from 
Havana. (Cape Cod 42° 2' N"., 70° 3' W. ; Havana, 23° 9' N., 
82° 22' W.) 

121. Leaving latitude 49° 57' N., longitude 15° 16' W., a 
ship sails between S. and W. till the departure is 194 miles, 
and the latitude is 47° 18' N. Find the course, distance, and 
longitude reached. 

122. Leaving latitude 42° 30' N., longitude 58° 51' W., a 
ship sails S.E. by S. 300 miles. Find the position reached. 

123. Leaving latitude 49° 57' K, longitude 30° W., a ship 
sails S. 39° W., and reaches latitude 47° 44' K Find the 
distance, and longitude reached. 

124. Leaving latitude 37° X., longitude 32° 16' W., a ship 
sails between N. and W. 300 miles, and reaches latitude 41° N. 
Find the course, and longitude reached. 

125. Leaving latitude 50° 10' S., longitude 30° E., a ship 
sails E.S.E., making a departure of 160 miles. Find the dis- 
tance, and position reached. 

126. Leaving latitude 49° 30' K, longitude 25° W., a ship 
sails between S. and E. 215 miles, making a departure of 167 
miles. Find the course, and position reached. 

127. Leaving latitude 43° S., longitude 21° W., a ship, sails 
273 miles, and reaches latitude 40° 17' S. What are the two 
courses and longitudes which will satisfy the data ? 

128. Leaving latitude 17° N., longitude 119° E., a ship sails 
219 miles, making a departure of 162 miles. What four sets 
of answers do we get ? 

129. A ship in latitude 30° sails due east 360 statute miles. 
What is the shortest distance from the point left to the point 
reached ? Solve the same problem for latitude 45°, 60°. 



106 



PLANE TRIGONOMETRY 



TRAVERSE SAILING 




Fig. 73 



Traverse Sailing is the application of 
the principles of Plane and Middle Lati- 
tude Sailing to cases when the ship sails 
from one point to another on two or 
more different courses. Each course is 
worked by itself, and these independent 
results are combined, as may be seen in 
the solution of the following examples. 



130. Leaving latitude 37° 16' S., longitude 18° 42' W., a 
ship sails N.E. 104 miles, then N.N.W. 60 miles, then W. by 
S. 216 miles. Find the position reached, and its bearing and 
distance from the point left. 

We have, for the first course, difference of latitude 73.5 N., 
departure 73.5 E. ; for the second course, difference of latitude 
55.4 N., departure 23 W. ; for the third course, difference of 
latitude 42.1 S., departure 211.8 W. 

On the whole, then, the ship has made 128.9 miles of north 
latitude, and 42.1 miles of south latitude. The place reached 
is therefore on a parallel of latitude 86.8 miles to the north of 
the parallel left, that is, in latitude 35° 49.2' S. 

The departure is, in the same way, found to be 161.3 miles 
W. ; and the middle latitude is 36° 32.6'. With these data, 
and the formula after Example 118, we find the difference 
of longitude to be 201', or 3° 21' W. Hence, the longitude 
reached is 22° 3' W. 

With the difference of latitude 86.8 miles, and the departure 
161.3 miles, we find the course to be K 61° 43' W., and the 
distance 183.2 miles. The ship has reached the same point 
that it would have reached if it had sailed directly on a 
course K 61° 43' W. for a distance of 183.2 miles. 



MISCELLANEOUS EXAMPLES 107 

131. A ship leaves Cape Cod (Example 120), and sails S.E. 
by S. 114 miles, N. by E. 94 miles, W.N. W. 42 miles. Solve 
as in Example 130. 

132. A ship leaves Cape of Good Hope (latitude 34° 22' S., 
longitude 18° 30' E.), and sails N.W. 126 miles, N. by E. 84 
miles, W.S. W. 217 miles. Solve as in Example 130. 

EXERCISE XXTV 
PROBLEMS IN GONIOMETRY 

Prove that : 

1. sin x + cos x — V2 cos (x — \ it). 

2. sin x — cos x = — v2 cos (x + i it). 

3. sin x + V3 cos x = 2 sin (x + ^ 7r). 

4. sin (x -\- ^ 7r) + sin (x — ^ 7r) = sin a;. 

5. cos (x + \ 7r) + cos (x — i. 7r) = V3 cos x. 

6. tan a; + sec x = tan (£ » + J *w)« 

7. ta n , + seo,= * 



8. 



sec a? — tan a? 
1 — tan x cot a; — 1 



1 -f tan x cot x + 1 



sin a? 1 + cos a; 

9. + : = 2 csc x. 

1 -f cos a? sm x 

10. tan a: + cot x = 2 csc 2 a:. 

11. cot x — tan as = 2 cot 2 a?. 

12. 1 + tan x tan 2 x = sec 2 #. 

sec 2 x 

13. sec 2 a; =7; — 

z — sec 2 a? 

14. 2 sec 2 x = sec (a; 4- 45°) sec (a-, - 45°). 



108 PLANE TRIGONOMETRY 

cos x + sin x 



15. tan 2 x + sec 2 x = 

16. sin2o5 = 



cos cc — sin cc 
2 tancc 



17. 2 sin a; + sin 2 cc 

18. sin 3 x = 

19. tan 3 x = 



1 + tan 2 x 

2 sin 3 x 



1 — COS X 

sin 2 2 x — sin 2 # 
sin ic 

3 tansc — tan 3 a: 



20 



1 — 3 tan 2 x 
tan 2 x + tan 05 sin 3 .r 



tan 2 jc — tan x sin o: 

21. sin (x + y) + cos (x — y) = 2 sin (a? -j- 1 7r) sin (y + i 7r). 

22. sin (05 + y) — cos (05 — y) = — 2 sin (x — ^ w) sin (y — i 7r). 

, J sin (05 + y) 

23. tan aj + tan v = — L ^ z * 

cos 05 cos y 

, , . sin 2 05 + sin 2 y 

24. tan (a; + 2/) = j—J- ;r*. 

v OJ cos 2 05 + cos 2 y 

25 sin o; + cos y = tan \\{x + y) + 45°j 
sin 05 — cos 3/ tan \\ {x — y) — 45° J 

26. sin 2 05 + sin 4 05 = 2 sin 3 05 cos 05. 

27. sin 4 05 — 4 sin o; cos 05 — 8 sin 3 o; cos 05 

= 8 cos 3 05 sin 05 — 4 cos 05 sin 05. 

28. cos 4 05 = 1 — 8 cos 2 05 + 8cos 4 o5 = 1 — 8 sin 2 05 + 8 sin 4 05. 

29. cos 2 03 + cos 4 05 = 2 cos 3 05 cos o\ 

30. sin 3 05 — sin x = 2 cos 2 or sin 05. 

31. sin 3 05 sin 3 05 + cos 3 o; cos 3 05 = cos 3 2 05. 

32. cos 4 05 — sin 4 05 = cos 2o5. 



MISCELLANEOUS EXAMPLES 109 



33. cos 4 x + sin 4 a: = 1 — \ sin 2 2 x. 

34. cos 6 cc — sin 6 a: = (1 — sin 2 ;*: cos 2 a-) cos 2 x. 

35. cos 6 a; 4- sin 6 a: = 1 — 3 sin 2 a- cos 2 x. 

sin 3 x -f- sin 5 a: 

36. — - = cot x. 

cos 6 x — cos 5 a: 

sin 3 a + sin 5 a 

37. — : : — - — — 2 cos 2 x. 

sin x -f- sin rf x 

38. esc x — 2 cot 2 a: cos a = 2 sin a. 

39. (sin 2 jc — sin 2 y) tan (jc + y) = 2 (sin 2 a- — sin 2 >/). 

sec x esc x 



esc- a; sec'a: 
sin 2 3a; 



40. (1 + cot a + tan a) (sin a — cos x) = 

41. sin x + sin 3 a 4- sin 5x = 

sin x 

3 cos x + cos 3 x . 

42. — — ; ; — = COt 8 X. 

6 sm x — sin 6 x 

43. sin 3 x = 4 sin a sin (6*0° + a) sin (60° — x). 

44. sin 4 x = 2 sin a: cos 3 x + sin 2 a\ 

45. sin # + sin (x — -J 7r) + sin (£ 7r — x) = 0. 

46. cos a: sin (?/ — g) 4- cos y sin (g — a-) 4- cos z sin (x — y)= 0. 

47. cos (a + y) sin ?/ — cos (x + *) sin g 

= sin (a- + ?/) cos y — sin (a* + z) cos g. 

48. cos (x + ?/ + g) + cos (a- + y — z) 4- cos (x — ?/ +• g) 

4- cos (y + g — x) = 4 cos x cos ?/ cos g. 

49. sin (a; 4- y) cos (x — y) 4- sin (?/ 4- g) cos (y — z) 

4- sin (g 4- a-) cos (g — a*) = sin 2 x 4- sin 2 y + sin 2 g. 

KA sin 75° + sin 15° „ AO 

50. - — — : — — — = tan 60 . 

sm 75° - sm 15° 

51. cos 20° 4 cos 100° + cos 140° = 0. 



110 PLANE TRIGONOMETRY 

52. cos 36° + sin 36° = V2 cos 9°. 

53. tan 11° 15' + 2 tan 22° 30' + 4 tan 45° = cot 11° 15'. 

If A, B, C are the angles of a plane triangle, prove that : 

54. sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C. 

55. cos 2 J. + cos 2 jB -f cos 2 C = — 1 — 4 cos J. cos 5 cos C 

56. sin 3 J. + sin 3 B + sin 3 C = — 4 cos -— cos — cos -jj— 

L Z Z 

57. cos 2 J. 4- cos 2 i? + cos 2 C = 1 — 2 cos A cos i? cos C 

If A +B + C = 90°, prove that : 

58. tan A tan E 4- tan B tan C 4- tan C tan 4=1. 

59. sin 2 J. + sin 2 £ -(- sin 2 C = 1 — 2 sin A sin 5 sin C. 

60. sin 2 4 4- sin 2 £ 4- sin 2 C = 4 cos /I cos B cos C. 

Prove that : 

61. sin (sin - ^ 4- sin -1 ?/) = x Vl — y % + y Vl — x 2 . 

3" 4" V 

62. tan (tan -1 re 4- tan -1 y) = — — • 

v 7 1 — x y 

63. 2 tan -1 x = tan -1 



l-x< 



64. 2 sin -1 a = sin -1 (2cc Vl — x 2 ). 

65. 2cos -1 z = cos -1 (2:r 2 — 1). 

66. 3 tan -1 # — tan -1 



67. sin 



-^ = tan -1 V 



3x- 


-a: 3 


1- 


3z 2 


V: 


— X 



, x — y . x — y 

68. sin -1 \/ ^ = tan -1 \ *• 

* X — z * y — z 



69. sin- 1 ^ = sec 



MISCELLANEOUS EXAMPLES 111 

1 



Vi 



, 2 Vx 2 
70. 2 sec -1 x = tan -1 - 



2 - x l 

71. tan" 1 ^ + tan- 1 ^ = 45°. 

72. tan- 1 ^ + tan" 1 ^ = tan" 1 *. 

73. sin" 1 ! + sin- 1 ^ = s i n -ig|. 

1 4 

74. sin -1 — -= + sin -1 — ■?= = 45°. 

V82 V41 

75. sec" 1 « + sec- 1 ! | = 75° 45'. 

76. tan- 1 (2 + V3)- tan" 1 (2 - V3) = sec~ 1 2. 

77. tan" 1 ^ -f tan" 1 J + tan" 1 ! + tan" 1 ^ = 45°. 

78. tan -1 :; - — - + tan -1 — - — — - = tan~ 1 7— -■ 

1 — 2^ + 4 x 2 1 + 2 x -f 4 a: 2 2x 2 

79. Given cos x = f ; find sin % x and cos •£- x. 

80. Given tan x = £ ; find tan ^ x. 

81. Given sin x + cos x = V^ ; find cos 2 x. 

82. Given tan 2x = - 2 T 4 - ; find sin x. 

83. Given cos 3 x = f f • find tan x. 

84. Given 2 esc x — cot x = V3 ; find sin ^ x. 

85. Find sin 18° and cos 36°. 

Find the value of : 

86. a sec x + b esc x, when tan x = \/— • 



87. sin 3 x, when sin 2x = Vl — ra 2 . 

88. sin x, when tan 2 x + 3 cot 2 a; = 4. 



112 PLANE TRIGONOMETRY 

csc 2 cc — sec 2 x . r- 

89. — — ? when tan x = Vi 

csc^sc 4- see^ce 7 

90. cos x, when 5 tan or + sec x = 5. 

91. sec x, when tan x = 



V2a + 1 

Simplify the following expressions : 

(cos x + cos y) 2 + (sin x -f sin ?/) 2 
cos 2 ^(^ — y) 

sin (# + 2 y) — 2 sin (ic + y) + sin cc 
cos (x -\-2y)— 2 cos (cc -f y) + cos cc 

sin (x — z) -\- 2 sin cc + sin (x + 2) 
sin (y — z) + 2 sin y -f- sin (2/ + 2) 

cos 6 x — cos 4 a? 

95. - — : — — • 

sin b # + sin 4 # 

96. tan- 1 (2x + l) + tan- 1 (2cc-l). 

Ill 



1 + sin 2 # 1 + cos 2 cc l + sec 2 ic 1 + csc 2 z 
98. 2 sec 2 # — sec 4 ;z — 2 csc 2 cc -f csc 4 ic. 

SOLUTION OF SINGLE EQUATIONS 

To solve a single equation that involves different functions 
of the same angle, or the same or different functions of related 
angles, first transform the equation, if necessary, into an 
equivalent equation that involves a single function of the 
same angle. 

Employ the method of factoring, if possible, in the algebraic 
part of the solution. 

Completely solve each equation, and check the results by 
substitution in the given equation. 



MISCELLANEOUS EXAMPLES 113 

Solve cos x = sin 2 as. 

By [12], p. 58, sin2x = 2 sin xcosx. 

.-. cos x = 2 sin x cos x. 
.*. (1—2 sin x) cos x = 0. 

.-. cos x = 0, or 1 — 2 sin x = 0. 
.-. x = 90° or 270°, or 30° or 150°. 
Each of these values satisfies the given equation. 

Solve the following equations : 

99. sin as = 2 sin(-^7r + x). 111. sin x = cos 2 x. 

100. sin 2 x = 2 cos x. 112. tan x tan 2 x = 2. 

101. cos 2 as = 2 sin x. 113. sec as = 4 esc a\ 

102. sin as + cos a: = 1. 114. cos + cos 2 6 = 0. 

103. sin as + cos 2 as = 4 sin 2 a:. 115. cot \0 + esc = 2. 

104. 4 cos 2 x + 3 cos as = 1. 116. cot as tan 2 a- = 3. 

105. sin x -f sin 2 as = sin 3 x. 117. sin as sec 2 x = 1. 

106. sin 2 as = 3 sin 2 as — cos 2 as. 118. sin 2 as + sin 2 as = 1. 

107. cot = ^tan 0. 119. cos as sin 2 as esc as = 1 . 

108. 2 sin = cos 0. 120. cot x tan 2 as = sec 2 x. 

109. 2 sin 2 as + 5 sin as = 3. 121. sin 2 as = cos 4 as. 

110. tan a* sec as = V2. 122. sin 2z cot 2 — sin 2 £=:i 

123. tan x -f- tan 2 as = tan 3 x. 

124. cot as — tan x = sin x + cos as. 

125. tan 2 as = sin 2 a*. 

126. tan x + cot as = tan 2 as. 

1 — tan a- _ 

127. = cos 2 x. 

1 + tan as 



114 PLANE TRIGONOMETRY 

128. sin x + sin 2 x = 1 — cos 2 x. 

129. sec 2 ic + 1 = 2 cos as. 

130. tan 2 a; + tan 3 a; = 0. 

131. tan (£ 7r + x) + tan (£ 7r — x) = 4t. 



132. Vl + sin x — Vl — sin x = 2 cos a:. 

133. tan aj tan 3 a? = — f. 

134. sin (45° + x) + cos (45° — x) = 1. 

135. tan a: + sec a; = a. 

136. cos 2 x = a (1 — cos x). 

137. (1 — tan x) cos 2 a: = a (1 + tan x). 

138. sin 6 a: + cos 6 a: = T 7 ^ sin 2 2 a:. 

139. cos 3 x + 8 cos 3 a; = 0. 

140. sec (a; + 120°) + sec (x - 120°) = 2 cos x. 

141. csc x = cot x 4- V3. 

142. 4 cos 2 x 4- 6 sin a? = 5 

143. cos x — cos 2x = 1. 

144. sin 4 a; — sin 2 a; = sin x. 

145. 2 sin 2 a3 4- sin 2 2 x = 2. 

146. cos 5 a; 4- cos 3 a; 4- cos a; = 0. 

147. sec x — cot a; = csc x — tan x. 

148. tan 2 a; 4- cot 2 a; = -*/. 

149. sin 4 a? — cos 3 a? = sin 2 as. 

150. sin x 4- cos a; = sec x. 

151. 2 cos x cos 3 a; 4- 1 — 0. 

152. cos 3 x — 2 cos 2 a; 4- cos x = 0. 



MISCELLANEOUS EXAMPLES 115 

153. tan 2x tana* = 1. 

154. sin (a; + 12°)+ sin (x - 8°) = sin 20°. 

155. tan (60° + x) tan (60° - x) = - 2. 

156. sin (z + 120°) + sin (z + 60°) = \. 

157. sin (x + 30°) sin (x - 30°) = J. 

158. sin 4 x 4- cos 4 a- = f . 

159. sin 4 a; — cos 4 a: = -g-g. 

160. tan (x + 30°) = 2 cos x. 

161. sec a; = 2 tan x + i. 

162. sin 11 x sin 4 x + sin 5 a; sin 2 a: = 0. 

163. cos cc + cos 3 x + cos 5 x + cos 7 x = 0. 

164. sin (x + 12°) cos (a; - 12°) = cos 33° sin 57°. 

165. sin- 1 «4-sin- 1 Jaj = 120°. 

166. tan~ 1 a; + tan- 1 2a; = tan- 1 3 V3. 

167. sin _1 x -f- 2 cos" \x = §7r. 

168. sin- 1 x + 3cos- 1 x = 210°. 

169. tan" 1 a; + 2 cot" 1 a; = 135°. 

170. tan" 1 (a + 1) + tan" 1 (a: — 1) = tan -1 2 a:. 

,x + 2 % x — 2 

171. tan" 1 H + tan" 1 - 3 - 



* + l ' a--l 4 

172. tan" 1 -^-5 =60°. 

1 — ar 

173. cos 2 (9 sec + sec + 1 = 0. 

174. sin x cos 2 a: tan x cot 2 a; sec x esc 2x = 1. 

175. sin i- a; (cos 2 a- - 2) (1 - tan 2 a-) = 0. 

Hint. Equate to each factor except the second. The second factor 
cannot equal 0. 



116 PLANE TRIGONOMETRY 

176. sin 3x = cos2x — 1. 178. sin 2 = cos 3 0. 

177. tan# + tan*2£c = 0. 179. (3-4cos 2 z)sin2a; = 0. 

180. sin x + sin 2x + sin 3 x = 0. 

181. sin0 + 2sin2 + 3sin30:=O. 

182. sin 2 # cos 2 # — cos 2 # — sin 2 ic + 1 = 0. 

183. sin x + sin 3 x = cos cc — cos 3 x. 



184. (1 — VI — tan 2 a) cos 2 a vers 3 x = 0. 

185. tan (0 -f 45°) == 8 tan 0. 

186. sin (aj — 30°) = £ V3 sin x. 

187. tan (0 + 45°) tan = 2. 



188. sin-^a^SO . 



SYSTEMS OF EQUATIONS 

189. Solve for x and y the system 

x sin a H- y sin ft = a, (1) 

x cos a -\- y cos f$ = b. (2) 

(1) x cos a, x sin a cos <:r + y sin /3 cos a = a cos a. (3) 

(2) x sin a, x sin a cos a + y cos /3 sin a = b sin a:. (4) 

(3) — (4), y (sin /3 cos a — cos /3 sin a) = a cos a — b sin a. (5) 

a cos a — 6 sin a 



••• 2/ 
Similarly, x 



sin (j3 — a) 

b Bin j8 — a cos /3 

sin (j3 — a) 



190. Solve for x and y the system 

sin x + sin y = a, (1) 

cos # + cos y = J. (2) 



MISCELLANEOUS EXAMPLES 117 

Transform (1) and (2), by Sect. XXXII, 

by [20], p. 59, 2 sin $(x + y) cos l(x-y) = a 1 (3) 

by [22], p. 59, 2 cos£(x + y) cos£(x - y) = b. (4) 

(3)-*-(4), tani(x + </) = ^. (5) 







• *3"* i \* t y; — 

Va 2 + b 2 


\ v ) 


Substitute value of 


sin£(z + y) in (3), 








cos \ (x — y) — £ Va- + 6 2 . 


(7) 


From (5), 




x + y — 2 tan— » - . 


(8) 


From (7), 


X 

y 


x -y = 2 cos- 1 ! Va 2 + ^ 2 . 


(9) 


Whence 


= tan- > - + cos- ! \ Va 2 + B 2 , 
6 




d 


= tan- 1 cos- 1 ! Va 2 + b 2 . 

b 




191. Solve for r 


and the system 








r sin = a, 


(1) 






r cos = J. 


(2) 


(1) + (2), 




tan 6 = - . 
b 


(3) 


From (3), 




^ = tan- 1 -. 
h 


(4) 


Square (1) 


and (2) and add, 

r 2 (sin 2 6 + cos 2 6) = a 2 4- b 2 . 





.-. r = 'v^i 2 + b 2 . 

192. Solve for r and the system 

r sin (0 + a) = a, (1) 

r cos (0 -f (3) = b. (2) 

Expand (1) and (2), 

r sin cos a: + r cos sin a = a. (3) 

r cos cos /3 — r sin sin /3 = 6. (4) 

Now solve (3) and (4) for r sin and r cos 0, as in Example 189. Then 
solve for r and 0, as in Example 191. 



118 PLANE TRIGONOMETRY 

193. Solve for r, 0, and <£ the system 



r cos <f> sin 6 = a, 


(1) 


r cos <f> cos = b, 


(2) 


r sin <£ — c. 


(3) 


(l)-4-(2), tan0 = -. .-. = tan~i a . 


(4) 


Square (1) and (2) and add, 

r 2 cos 2 tf> = a 2 + & 2 . 


(5) 


(3) • (5) tan <p — ■ • .-. <f> — tan -1 


C -(6) 

+ 62 


Va 2 + & 2 Va 2 
Square (3) and add to (5), 

r 2 = a 2 + & 2 + c 2 . 



.-. r = Va 2 + 6 2 + c 2 . (7) 

Solve the following systems for r, 0, cf>, x, and y : 

194. x sin 21° + y cos 44° = 179.70, 
a; cos 21° + y sin 44° = 232.30. 

195. sin x — sin y = 0.7038, 
cos x — cos y = — 0.7245. 

196. r sin 6 = 92.344, 
r cos = 205.309. 

197. r sin (0-19° 18') = 59.4034, 
r cos (0 - 30° 54') = 147.9347. 

198. r cos <£ cos = — 46.7654, 
r sin <f> cos 6 = 81, 

r sin = — 54. 

199. Eliminate from the system 

x = r (0 — sin 0), 

y = r(l — COS 0). 
Hint. 1 — cos = vers 0. .-. = vers -1 - • 



CHAPTER VI 

CONSTRUCTION OF TABLES 

SECTION" XLIII 

LOGARITHMS 

Properties of Logarithms. Any positive number except unity- 
being selected as a base, the index or exponent which the base 
must have to produce a given number is the logarithm of that 
number to the given base. 

Thus, if a n = N, then n = log a iV. 

n = log a iV is read, n is equal to log N to the base a. 
Let a be the base, M and N any positive numbers, m and n 
their logarithms to the base a ; so that 
a m = M, a n = X, 

m = log a M, n = \og a N. 

Then, in any system of logarithms : 

1. The logarithm of 1 is 0. 

For, a° = l. .'. = log l. 

2. The logarithm of the base itself is 1. 
For, a 1 = a. .'. 1 = log a a. 

3. The logarithm of the reciprocal of a positive number is 
the negative of the logarithm of the number. 

For, if a n = N, then — = — = a _n . 

' iV a n 

.*• log a (-J = -n = - log a N. 
119 



120 PLANE TRIGONOMETRY 

4. The logarithm of the product of two or more positive 
numbers is found by adding together the logarithms of the 
several factors. 

For, ' M X N = a m x a n = a m + n . 

.*. log a (M xN) = m + n = log a M -f log a N. 
Similarly for the product of three or more factors. 

5. The logarithm of the quotient of two positive numbers 

is found by subtracting the logarithm of the divisor from the 

logarithm of the dividend. 

^ M a m 

For, - = - = *—., 



% 



• •'• loga l ]v J = m ~ n = lo g«^ - log a N. 

6. The logarithm of a power of a positive number is found 
by multiplying the logarithm of the number by the exponent 
of the power. 

For, N p = (a n ) p = a np . 

.-. \og a (N p ) = np =p log a N. 

7. The logarithm, of the real positive value of a root of a 
positive number is found by dividing the logarithm of the 
number by the index of the root. 

For, -Vn = ->/a" = a?. 

r r 

Change of System. Logarithms to any base a may be con- 
verted into logarithms to any other base b as follows : 
Let N be any number, and let 

n = log a iV and m = log 6 iV~. 
Then, N = a n and N = b m . 

.'. a n = b m . 



CONSTRUCTION OF TABLES 121 

Taking logarithms to any base whatever, 
n log a = m log b, 
or, log a x log tt iV = log b x log 6 iV, 

from which log b N may be found when log a, log b, and log a .V 
are given; and conversely, log a iV may be found when log a, 
log b, and log 6 iV are given. 

Two Important Systems. Although the number of different 
systems of logarithms is unlimited, there are but two systems 
which are in common use. These are : 

1. The common system, also called the Briggs, denary, or 
decimal system, of which the base is 10. 

2. The natural system of which the base is the fixed value, 
which the sum of the series 

1 + 1 + L2 + 12^3 + 1.2.3.4 + ' " 

approaches as the number of terms is indefinitely increased. 
This fixed value, correct to seven places of decimals, is 
2.7182818, and is denoted by the letter e. 

The common system is used in actual calculation; the 
natural system is used in the higher mathematics. 

EXERCISE XXV 

1. Given log 10 2 = 0.30103, log 10 3 =0.47712, log 10 7 =0.84510; 
find log 10 6, log 10 14, log 10 21, log 10 4, log 10 l2, log 10 5, log 10 £, 

logwj, logioi, log 10 fl. 

2. With the data of Example 1, find log 2 10, log 2 5, log 3 5, 
log 7 £, logg^. 

3. Given log 10 e = 0.43429 ; find log e 2, log e 3, log, 5, log e 7, 
log e 8, log, 9, log e f, log e f, log, |f, log el5 -V 

4. Find x from the equations 5* = 12, 16* = 10, 27* = 4. 



122 PLANE TRIGONOMETRY 

SECTION XLIV 
EXPONENTIAL AND LOGARITHMIC SERIES 
Exponential Series. By the binomial theorem, 



H) 



. , 1 , nx (nx — 1) 1 

l + nxX- + — Y"^ x 1 

n 1-2 n z 

, Tio? (nx — 1) (was — 2) 1 
H — X h • 

^ 1.2-3 rc 8 



( 1\ ( 1\( 2\ 
x\ x J x\ x — - \\ x 1 

= 1 + , + A^ + V »A ^ + .„ (1) 

This equation is true for all real values of x, since the bino- 
mial theorem may be extended to the case of incommensurable 
exponents (Wentworth's College Algebra, § 299) ; it is, how- 
ever, true only for values of n numerically greater than 1, since 

- must be numerically less than 1 (College Algebra, § 418). 

As (1) is true for all values of x, it is true when x = 1. 



b„, [K-)]-H) _ 

Hence, from (1) and (2), 

12 |3 + 

= i + x + ^_^ + ^ 1\ /+. 



CONSTRUCTION OF TABLES 



123 



This last equation is true for all values of n numerically 
greater than 1. Taking the limits of the two members as n 
increases without limit, we obtain 



1+1+ | + ^ + 



= 1 +*+f+f+">< 3 } 



and this is true for all values of x. It is easily seen that both 
series are convergent for all values of x. 

The sum of the infinite series in parenthesis is the natural 
base e. 



Hence, by (3), e* = 1 + x + % + ,|- + . . • 

To calculate the value of e, we proceed as follows : 

1.000000 



1.000000 



0.500000 



0.166667 



0.041667 



0.008333 



0.001388 



0.000198 



0.000025 



Adding, 

To ten places, 



Limit of 1 + 



H) 



0.000003 

e = 2.71828. 

e = 2.7182818284. 

By the binomial theorem, 



1+nX-l 
n 



x t n (n — 1) x 2 
rr 



= l + z + 



12 

n(n-l)(n-2) x* 
+ 1.2-3 X n*^' 

n „* , \ n )\ n ) 



(4) 



[2 



L5 



x* + 



124 PLANE TRIGONOMETRY 

This equation is true for all values of n greater than x 
{College Algebra, § 418). Take the limit as n increases with- 
out limit, x remaining finite ; then 

JSi( 1 +=)'- 1 +-+|+|+- 

Logarithmic Series. 

Let y = log c (l+a;); 



(?) 



then 1 + x = & = n 1[ ? lit ( 1 + | 



If n is merely a large number, but not infinite, 



c 



where e is a variable number which approaches the limit 0, 
when n increases without limit. Hence, 

1 + | = Vl+* + e, 

y = n VI + cc + e — n. 

If ?i increases without limit, and consequently e approaches 
as a limit, we have 

limit 



y = lmut \n<Sl + x- n\ 

If x is less than 1, we may expand the right-hand member 
of this equation by the binomial theorem. The result is 

.-.■ttt-H-KHpt-"}-] 

^KHi+e- r XHg-] 

x i \2x 3 \Sx 4 



CONSTRUCTION OF TABLES 125 



ar x* x 



.-Aog e (l + x) = x -_ + --- + ... 

This series is known as the logarithmic series. It is con- 
vergent only if x lies between — 1 and + 1, or is equal to + 1. 
Even within these limits it converges rather slowly, and for 
these reasons it is not well adapted to the computation of loga- 
rithms. A more convenient series is obtained as follows. 



Calculation of Logarithms. The equation 

log.(l + y) = y-| + | 3 -^ + --- (1) 



holds true for all values of y numerically less than 1 ; there- 
fore, if it holds true for any particular value of y less than 1, 
it will hold true when we put — y for y\ this gives 

icg.(i-y) = -y-|-|'- '{-■■ (2) 

Subtracting (2) from (1), since 

log, (l + z,) -log, (1-2,) = log/ 

weftnd hfc («±l)_,(, + 5 + J 



Put 
then 



2« + l' 

1 + .', •- + 1 






and log e f ^-;— j = log«(« + l)-log«« 

V2« + 1^3(2« + l) 3 + 5(2s + l) 5 + 
This series is convergent for all positive values of z. 



■> 



126 



PLANE TRIGONOMETRY 



Logarithms to any base a can be calculated by the series : 
log a (z + l)-log £ 

1 1 



—LL(-L- 



+ 



3(2z + l) 8 5(2z + l) 



+ 



■> 



Example. Calculate log c 2 to five places of decimals. 
Let z = 1 ; then 2 + 1 = 2, 2z + 1=3, 



and 



log, 2 



? + __*_ 

3 3 x 3 1 



+ 



-.7.+ 



2 



5 x 3 5 7x3' 



+ 



The work may be arranged as follows : 



2 .000000 

0.660667 -s- 1 = 0.666667 

0.074074 -=- 3 = 0.024691 

0.008230 -T- 5 = 0.001646 

0.0 00914 -=- 7 = 0.000131 

000102 -j- 9 = 0.000011 
0.000011 - 11 = 0.000001 



log e 2 =0.693147 



Note. In calculating logarithms the accuracy of the work may be 
tested every time we come to a composite number by adding the logarithms 
of the several factors. In fact, the logarithms of composite numbers are 
best found by addition, and then only the logarithms of prime numbers 
need be computed by the series. 



EXERCISE XXVI 

1. Calculate to five places of decimals log e 3. 

2. Calculate to five places of decimals log e 5. 

3. Calculate to five places of decimals log e 7. 

4. Calculate to ten places of decimals log e 10. 

5. Calculate to five places of decimals log 10 2, log 10 e, log 10 ll. 



CONSTRUCTION OF TABLES 



127 



SECTION XLV 



TRIGONOMETRIC FUNCTIONS OF SMALL ANGLES 



Let A OP (Fig. 74) be any angle less than 90° and x its cir- 
cular measure. Describe a 
circle of unit radius about 
as a centre and take 
ZAOP' = -ZAOP. Draw 
the tangents to the circle at 
P and P', meeting OA in T. 
Then, from Geometry, 

chord PP' < arc PP' 

<PT + P'T, 
or, by dividing by 2, FlCr . 74 




or 



ikfP < arc .4P<PT, 
sin x < x < tan x. 



Hence, dividing by sin x, 



1 < — < sec x, 



sin x 



sin a; 
1 > > cos x. 



(i) 



Then lies between cos x and 1. 



If now the angle x is constantly diminished, cos x approaches 
the value 1. 



Accordingly, the limit of 



smx 



as x approaches 0, is 1. 



In other words, if x is a very small angle, then differs 

from 1 by a small value c ; and this small value c approaches 
as a; approaches 0. 



128 PLANE TRIGONOMETRY 

Example. To find the sine and cosine of 1'. 
If x is the circular measure of 1', 

the next figure in x being 8. 

Now sincc>0 but < x ; hence, sin 1' lies between and 
0.000290889. 



Again, cos 1' = Vl - sin 2 l' > Vl - (0.0003) 2 > 0.9999999. 
Hence, cos V = 0.9999999+. 

But, from (1), sin x > x cos x. 

.*. sin l'> 0.000290888 x 0.9999999 

> 0.000290888 (1 - 0.0000001) 

> 0.000290888 - 0.0000000000290888 

> 0.000290887. 

Hence, sin 1' lies between 0.000290887 and 0.000290889; 
that is, to eight places of decimals 

sin 1' = 0.00029088+, 

the next figure being 7 or 8. 

EXERCISE XXVII 

Given 7T = 3.141592653589 : 

1. Compute sin 1', cos 1', and tan 1' to eleven places of 
decimals. 

2. Compute sin 2' by the same method, and also by the 
formula sin 2 x = 2 sin x cos x. Carry the operations to nine 
places of decimals. Do the two results agree ? 

3. Compute sin 1° to four places of decimals. 



CONSTRUCTION OF TABLES 129 

x 

4. From the formula cos x = 1 — 2 sin 2 -> show that 

Li 

COS X > 1 — — • 

5. Show by aid of a table of natural sines that sin x and x 
agree to four places of decimals for all angles less than 4° 40'. 

6. If the values of log x and log sin x agree to five decimal 
places, find from a table the greatest value x can have. 



SECTION XLVI 
SIMPSON'S METHOD OF CONSTRUCTING A TRIGONOMETRIC TABLE 

By Sect. XXXII, p. 59, 

sin (A + B) + sin (A - B) = 2 sin A cos B. 

If we put A = x + 2 y, B = y, 

we have sin (x + 3 y) + sin (x -f ?/) = 2 sin (a; + 2 y) cos ?/, 

or sin (as + 3 y) = 2 sin (x + 2y) cos ?/ — sin (x + ?/). 

Similarly, cos (# + 3 ?/) = 2 cos (jc + 2 ?/) cos ?/ — cos (x + y). (1) 
If y = l',.the last two equations become 

sin(z + 3')= 2sin(^ + 2 f )cosl' — sin (a: 4- 1'), 
cos (x + 3') = 2 cos (x 4- 2') cos 1' — cos (x + 1'). 

Hence, taking x successively equal to — 1', 0', 1', 2', • • •, we 

obtain 

sin 2' = 2 sin 1' cos 1', 

sin 3' = 2 sin 2 f cos V — sin 1', 

sin 4' = 2 sin 3' cos 1' - sin 2', 



cos 2' 


= 2cos 2 r-l, 




cos 3' 


= 2 cos 2' cos 1' 


— cos 1', 


cos 4' 


= 2 cos 3' cos 1' 


— cos 2 f , 



130 PLANE TRIGONOMETRY 

Since sin 1' and cos 1' are known, these equations enable us 
to compute step by step the sine and cosine of any angle. 
The tangent may then be found in each case as' the quotient 
of the sine divided by the cosine. 

This process need be carried only as far as 30°. For 

sin (30° + x) + sin (30° - x) = 2 sin 30° cos x = cos x, 
cos (30° + x) — cos (30° — x) = — 2 sin 30° sin x = — sin x. 
. ' . sin (30° + x) = cos x - sin (30° - x), 
cos (30° + x) = - sin x +■ cos (30° - x). 

Moreover, the sines and cosines need be calculated only to 
45°, since 

sin (45° + x)= cos (45° — x), 

cos (45° + x)= sin (45° — x). 

In using this method, the multiplication by cos 1', which 
occurs at each step, can be simplified by noting that 

cos V = 0.9999999 = 1 - 0.0000001. 

Note. Simpson's method is superseded in actual practice by much 
more rapid and convenient processes in which we employ the expansions 
of the trigonometric functions in infinite series. 



EXERCISE XXVm 

1. Compute the sine and cosine of 6' to seven decimal places. 

In Formula (1) let y = 1°. Assuming 

sin 1° = 0.01 7454+, cos 1° = 0.999848+ : 

2. Compute the sine and cosine of two degrees. 

3. Compute the sine and cosine of three degrees. 

4. Compute the sine and cosine of four degrees. 

5. Compute the sine and cosine of five degrees. 



CONSTRUCTION OF TABLES 131 

SECTION XLVII 

DE MOIVRE'S THEOREM 

Expressions of the form 

cos x + i sin x, 

when i = V— 1, play an important part in modem analysis. 
Given two such expressions, 

cos x -f- i sin x, cos y + * sin y 
their product is 

(cos x -f- i sin x) (cos y + t sin y) 

= cos a: cos y — sin a sin y + i (cos x sin y + sin x cos y) 
= cos (x -f y) + i sin (a; -f y). 

Hence, the product of two expressions of the form 

cos x + i sin x, cos y + i sin y 

is an expression of the same form in which x or y is replaced 
by x + y. In other words, the angle which enters into such a 
product is the sum of the angles of the factors. 

If x and y are equal, we have at once, from the preceding, 

(cos x + i sin x) 2 — cos 2 x -f- i sin 2 a; ; 
and again, 

(cos a: + i sin #) 8 = (cos x + i sin z) 2 (cos £ -+• * sin x) 

= (cos 2x + i sin 2 a:) (cos x + i sin a;) 
= cos 3 x + t sin 3 #. 
Similarly, 

(cos x + i sin a;) 4 = cos &x + i sin 4 a;, 
and in general, if n is a positive integer, 

(cos x + i sin a:) n = cos nx + i sin rac. (1) 



132 PLANE TRIGONOMETRY 

Hence, 

To raise the expression cos x + i sin x to the nth power when 
n is a positive integer, we have only to multiply the angle x by n. 

Again, if n is a positive integer as before, 

(x , . . x\* . . 

cos - + % sin - = cos x -f- i sin x. 
n nj 

. . . v- x . . X 

.'. (cos x + i sin x ) B = cos — t- i sin — 

v ' n n 

Since, however, x may be increased by any integral multiple 
of 2 it without changing cos x + i sin x, it follows that all the 
n expressions, 

x , . . x x-\-2tt , . . x-\-2tt 

cos — h % sin -? cos h i sin > 

?i w n n 

X + 4:7T . . X + 4 7T 

cos h t sm ' • • • > 

n n 

x-\- (n — 1)2 it . . x + (w — 1) 2 7r 

cos - h t sm j 

n n 

are nth roots of cos x + i sin #. There are no other roots, since 

x-\-n2ir , . . x-\-n27r 

cos h t sm 

n n 

( x i o "\ i • • ( x . o ^ # . a; 

= cos - + 2 7r J + i sm - + 2 7r = cos - + % sm - > 
\n J \n J n n 

x -f- (n + 1) 2 it , . . a; + (n + 1) 2 it 

and cos \- 1 sm * 

n n 

(x + 2ir \ . . fx+ 2tt \ 

= cos ( h 2 it J + i sm ( h 2 7r 1 

a; + 2ir , . . x -\-2tt 

= cos h i sm > 

w n 

and so on. 



CONSTRUCTION OF TABLES 133 

Hence, if n is a positive integer, 
(cos x + i sin x)" 

= cog £±2^ + . sin E±2*E (ft = , 1( 2 ) ...„_i ) . (2) 

From (1) and (2) it follows at once that if m and n are 
positive integers, 

m 1 

(cos x + i sin se) B = \ (cos a? + i sin x) n \ m 

iyl in 

= cos — (x + 2 &7r) + i sin — (a? + 2 &7r) 

7i X 7 11 J 

(k = 0, 1, 2, ••• 7i - 1). (3) 
Finally, if is a negative fraction, 

-- 1 

(cos x + i sin a?) " = 



But 



(cos x H- i sin a;)" 

cos x — i sin x 



cos x + i sin a? (cos x + i sin a") (cos x — i sin a:) 

_ cos x — i sin x 
cos 2 a; + sin 2 a; 

= cos x — i sin a; 

= cos (— x) -f- 1 sin (— a:). 

m m 

Hence, (cos x -f- i sin a;) n = J cos (— x) -f- i sin (— x) \ n 

7Yl> TTV 

— cos — (— x + 2 &7r) + » sin — (— x + 2 &7r), 

71 V 7 71 ' 

(A = 0, 1, 2, • • • n - 1) 

= cos \ (x + 2 &7r) I -M sin J (a; -f 2 &7r) I > 

(A = 0,1, 2,..- Ti-1). (4) 



134 PLANE TRIGONOMETRY 

Consequently, if n is a positive or negative integer or 
fraction, 

(cosx 4- * sinx) n = cos [n (x -+- 2 &7r)] + i sin [n. (x + 2 &7r)], 

(t = 0,l,2,-ji- 1). (5) 

Example. Find the three cube roots of — 1. 

We have - 1 = cos 180° + i sin 180°. 

, , x4 180° + 2 kit . . 180° + 2&7r, f A , nx 
.-. (- l)* = cos +isin (A; = 0, 1, 2). 

o o 

For the three cube roots of — 1 we find, therefore, 

cos 60° + i sin 60°, cos 180° -f i sin 180°, cos 300° + i sin 300°, 
1 + iV3 , 1 -iVs 



2 



-1, 



By aid of De Moivre's Theorem, we may express sin nO and 
cos nO, when n is an integer, in terms of sin and cos 0. 

Thus, cos nO -f- i sin nO = (cos + i sin 6) n 

= cos n + in cos"" 1 * sin 6 + i 2 ^ ~ ) C os n - 2 sin 2 

\A 

+ i . n(n-iyn-2) co8 ,., tfBin , g + v 

Or, since i 2 = — 1, i s = — i, i A = -f 1, • • •, 

cos rc0 -f- i sin rc0 = cos M -f in cos n_1 sin 6 

If. l£ 

Equating now the real parts and the imaginary parts sepa- 
rately, we obtain 

cos nO = cos n - [~ ' cos n ~ 2 sin 2 

+ — 5 ^-r- — ^ ^cos n - 4 0sm 4 , 



COXSTRUCTIOX OF TABLES 135 



, „ . n(n — 1) (n — 2) . _ . « 

sin ?*0 = ?i cos n_1 sm » v~ cos n-3 sm a 



I? 

n (n -l)(n-2)(?i-S)(?i -4) 



,_ cos' , ~ J sim 



EXERCISE XXIX 

1. Find the six 6th roots of — 1 ; of + 1. 

2. Find the three cube roots of i 

3. Find the four 4th roots of — i. 

4. Express sin 4 and cos 4 6 in terms of sin 6 and cos 0. 

SECTIOX XLVTII 

EXPANSION OF SIN X, COS X, AND TAN X IN INFINITE SERIES 

Let one radian be denoted simply by 1, and let 

cos 1 + i sin 1 = k. 
Then cos as -f- i sin x = (cos 1 -\- i sin l) x = l x . 

and, putting — a* for x, 

cos (— as) + i sin (— as) = cos .r — t sin x = fe~*. 
That is, cos as + £ sin as = A- r . 

and . cos as — i sin as = k~ x . 

By taking the sum and difference of these two equations, 
and dividing the sum by 2 and the difference by 2 /. we have 

1 1 

cos x = - (Jc x + k~% sin as = — (A~ r - k~ x ). 

But k x = (e l0 ? *) x = e* lo * *. &-* = e~* l0? * 

-, JlfW . - , , • , ^(logA-V , .r 3 flog Am 3 , 
and e**** = 1 -+- x log A- H r^ + r~— L H , 



136 PLANE TRIGONOMETRY 

e _*,„ g * = i _ x log k + ^WOf _ e!(M! + . . . 

i\ [3 [5 

It only remains to find the value of k, and this can be 
obtained by dividing the last equation through by x and 
letting x approach indefinitely. 

Then we have 



limit / sin a; 



H 



log k. 



But 



limit / sin x \ 
x -°\ x / 

. ' . log k = i. 

.'. k = e\ 
Therefore, we have 
1 



cos* = -(^+e-^)=l-| + g-j| 



-r77 + 



/ytti /^.O ™7 

From the last two series we obtain, by division, 

sin x , x 3 , 2 x 5 , 17 cc 7 

tana: == = a; -f — + -j^- + -^rz 

cos x 3 15 615 

By the aid of these series the trigonometric functions of any 
angle are readily calculated. 

In the computation it must be remembered that x is the 
circular measure of the given angle. 



CONSTRUCTION OF TABLES 137 

EXERCISE XXX 

Verify by the series just obtained that : 

1. sin 2 ic + cos 2 a? = 1. 

2. sin (— x) = — sin x and cos (— x) = cos x. 

3. sin 2 x — 2 sin sc cos cc. 

4. cos 2 a; = 1 — 2 sin 2 sc. 

5. Find the series for sec x as far as the term containing 
the Gth power of x. 

6. Find the series for x cot x, noting that 

x 

x cot x = — COS X. 

since 

7. Calculate sin 10° and cos 10° to five places of decimals. 

8. Calculate tan 15° to five places of decimals. 

9. From the exponential value of cos x show that 

cos 3 # = 4 cos 3 # — 3 cos x. 
10. From the exponential value of sin x show that 
sin 3 x = 3 sin x — 4 sin 3 ic. 



SPHERICAL TRIGONOMETRY 



CHAPTER VII 



THE RIGHT SPHERICAL TRIANGLE 



SECTION XLIX 
INTRODUCTION 



The object of Spherical Trigonometry is to explain the 
method of solving spherical triangles. To solve a spherical 
triangle is to compute any three of its parts when the other 
three parts are given. 





Fig. 75 



Fig. 76 



The sides of a spherical triangle are arcs of great circles. 
Thus, AB (Fig. 76) is an arc of a great circle. The sides of a 
spherical triangle are measured in degrees, minutes, and sec- 
onds, and therefore by the plane angles formed by radii of 

139 fc 



140 



SPHERICAL TRIGONOMETRY 



the sphere drawn to the vertices of the triangle. Therefore, 
the measures of the sides are independent of the length 
of the radius, which may be assumed to have any convenient 
numerical value ; as, for example, unity. 

The angles of a spherical triangle are measured by the 
dihedral angles made by the planes of the sides. Each angle 
is also measured by the number of degrees in the arc of a 
great circle, described from the vertex of the angle as a pole, 
and included between the sides of the angle. 

The sides may have any value from 0° to 360° ; but in this 
work only sides that are less than 180° will be considered. 
The angles may have any value from 0° to 180°. 

A right spherical triangle may have one, two, or three right 
angles. 

When a spherical triangle has one or 
more of its sides equal to a quadrant it is 
called a quadrantal triangle (Fig. 77). 

If any two parts of a spherical triangle 
are either both less than 90° or both greater 
than 90°, they are said to be alike in kind ; 
but if one part is less than 90°, and the 
other part greater than 90°, they are said to be unlike in kind. 




fig. 77 




Isosceles 



Equilateral 



Eight 



Oblique 



Fig. 78 



Spherical triangles are named isosceles, equilateral, equi- 
angular, right, and oblique, under the same conditions as plane 
triangles are named isosceles, equilateral, equiangular, right, 
and oblique. 



THE RIGHT SPHERICAL TRIANGLE 



141 



The following propositions are proved in Geometry. (See 
Wentworth's Geometry, §§ 815, 790, 795, 793.) 

1. If two angles of a spherical triangle are unequal, the 
sides opposite are unequal, and the greater side is opposite 
the greater angle ; and conversely. 

2. The sum of the sides of a spherical triangle is less 
than 360°. 





Fig. 79 



Fig. 80 



3. The sum of the angles of a spherical triangle is greater 
than 180° and less than 540°. 

4. If from, the vertices of a spherical triangle as poles, 
arcs of great circles are drawn, another triangle is formed so 
related to the first that each angle of either triangle is the 
supplement of the side opposite it in the other triangle. 





Fig. 81 



Fig. 




142 SPHERICAL TRIGONOMETRY 

Two spherical triangles, drawn as explained in Theorem 4, 
p. 141, are called polar triangles, or supplemental triangles. 

Let A, B, C (Fig. 83) denote the angles of one triangle ; 
a, b, e the sides opposite these angles, respectively ; and let 
A', B', C and a', b', c' denote the corresponding angles and 
the corresponding sides of the polar triangle. Then Theorem 4 
gives the six following equations : 

A + a' = 180°, 
B + V = 180°, 
C + c' = 180°, 
A' + a =180°, 
B' + b =180°, 
C' + c =180°. 



EXERCISE XXXI 

1. The angles of a triangle are 70°, 80°, and 100°. Find the 
sides of the polar triangle. 

2. The sides of a triangle are 40°, 90°, and 125°. Find the 
angles of the polar triangle. 

3. Show that, if a triangle has three right angles, the sides 
of the triangle are quadrants. 

4. Show that, if a triangle has two right angles, the sides 
opposite these angles are- quadrants, and the third angle is 
measured by the number of degrees in the opposite side. 

5. How can the sides of a spherical triangle, measured in 
degrees, be found in units of length, when the length of the 
radius of the sphere is known ? 

6. Find the lengths of the sides of the triangle in Example 2 
if the radius of the sphere is 4 feet. 



THE RIGHT SPHERICAL TRIANGLE 



143 



SECTION L 

FORMULAS RELATING TO RIGHT SPHERICAL TRIANGLES 

As is evident from Examples 3 and 4, Exercise XXXI, the 
only kind of right spherical triangle that requires further 
investigation is that which contains only one right angle. 





Fig. 84 



Fig. 85 



Let ABC (Fig. 85) be a right spherical triangle having only 
one right angle ; and let A, B, C denote the angles of the tri- 
angle ; a, b, c, respectively, the opposite sides. 

Let C be the right angle ; and for the present suppose that 
each of the other parts is less than 90°, and that the radius of 
the sphere is 1. 

Let planes be passed through the sides, intersecting in the 
radii OA, OB, and OC. 

Also, let a plane _L to OA be passed through B, cutting OA 
at E and OC at D. Draw BE, BD, and DE. 

BE and DE are each _L to OA (Wentworth's Geometry, § 501) ; 
therefore, Z BED = A. The plane BDE is _L to the plane A OC 
(Wentworth's Geometry, § 554) ; hence, BD, which is the inter- 
section of the planes BDE and BOC, is _L to the plane AOC 
(Wentworth's Geometry, § 556), and therefore _L to OC and DE. 



144 




SPHERICAL TRIGONOMETRY 

B Now, 

cos c = OE — ODx cos b, 
and OB = cos a. 
Therefore, 

cos c = cos a cos b. [38] 
Again, sin a = BB = BE x sin A, 
and BE = sin c. 



E 
Fig. 86 
Therefore, sin a = sin c sin A 



changing letters, sin b = sin c sin B 
Again, 



Hence, 
changing letters, cos B = tan a cot c J 



BE OE tan b 

cos A = — = — 

BE OE tan c 

cos A = tan b cot c 1 



[39] 



[40] 



Again, 



BE OB sm b 

cos A = -— — = : 

BE sm c 



= cos a 



sin ft 
sin c 



By substituting for 



sin 6 



sm c 

cos A = cos a sin B | 
changing letters, cos B = cos b sin A J 



its value from [39], we obtain 

[41] 



Again, 



. 7 BE BB cot A 
sm o = —— = —— — = tan a cot A. 



Hence, sin b = tan a cot A 

changing letters, sin a = tan b cot B 



[42] 



If in [38] we substitute for cos a and cos b their values 
from [41], we obtain 

cos c = cot A cot B. [43] 

Note. In order to deduce the second formulas in [39]-[42] geomet- 
rically, the auxiliary plane must be passed through A _L to OB. 



THE RIGHT SPHERICAL TRIANGLE 



145 



These ten formulas are sufficient for the solution of any 
right spherical triangle. In deducing these formulas, all the 
parts of the triangle, except the right angle, were assumed to 
be less than 90°. But the formulas hold when this hypothesis 
is not true. 

Let one of the legs a be greater than 90°, and construct a 
figure for this case (Fig. 88) in the same manner as Fig. 85. 





Fig. 87 



Fig. 88 



The auxiliary plane BDE will now cut both CO and AO 
produced beyond the centre O; and we have 

cos c = — OE = — OD cos DOE 

= — (— cos a) cos b — cos a cos b. 

Likewise, the other formulas, [39]-[43], hold true in this case. 
Again, suppose that both the legs a and b are greater than 
90°. In this case the plane BDE will cut CO produced beyond 
0, and A between A and ; and we have 

cos c — OE = OD cos DOE 

= ( — cos a) (— cos b) = cos a cos b, 

a result agreeing with [38]. 

Likewise the other formulas, [39]-[43], hold true in this case. 

Like results may be obtained in all cases. 

In other words, Formulas [38]-[43] are universally true. 



146 SPHERICAL TRIGONOMETRY 

EXERCISE XXXII 

1. Show, by aid of Formula [38], p. 144, that the hypote- 
nuse of a right spherical triangle is less than or greater than 
90°, according as the two legs are alike or unlike in kind. 

2. Show, by aid of Formula [41], that in a right spherical 
triangle each leg and the opposite angle are always alike in 
kind. 

3. What inferences may be drawn from Formulas [38]- 
[43] respecting the values of the other parts : (i) if e = 90° ; 
(ii) if a = 90° ; (iii) if c = 90° and a = 90° ; (iv) if .a = 90° 
and b = 90° ? 

Deduce from Formulas [38]-[43] and Formulas [18]-[23] 
the following formulas : 

4. tan 2 ib = tan -J (c — a) tan i(p -\- a). 

Hint. Substitute in Formula [18] the value of cos b from [38]. 

5. tan 2 (45° — % A ) = tan \ (c — a) cot i (c + a). 

6. tan 2 -j-f> = sin (c — a) esc (c + a). 

7. tan 2 i c = - cos (A + B) sec (A — B). 

8. tan 2 ^ a = "tan [-J- (4 + 5) - 45°] tan [-J- (4 - 5) + 45°]. 

9. tan 2 (45° - f c) = tan £ (^ - a) cot J (.4 + a). 

10. tan 2 (45° ~ib)= sin (4 — a) esc (4 + a). 

11. tan 2 (45 -i£) = tan£(>l -a) tan £(4 + a). 

SECTION LI 
NAPIER'S RULES 

The ten formulas deduced in Sect. L express the relations 
of five parts of a right triangle, the three sides and the 
two oblique angles. All these relations may be shown to 



THE RIGHT SPHERICAL TRIANGLE 147 

follow from two very useful rules, devised by Baron Napier, 
the inventor of logarithms. 

For this purpose the right angle (not entering the formulas) 
is not taken into account, and instead of the hypotenuse and the 
two oblique angles their respective complements are employed ; 
so that the five parts considered by Napier's Rules are : a, b, 
Co. A, Co. c, Co. B. 

Any one of these parts may be called a middle part ; and then 
the two parts immediately adjacent are called adjacent parts, 
and the other two are called opposite parts. Napier's Rules are 

Rule I. The sine of any middle part is equal to the product 
of the tangents of the adjacent parts. 

Rule II. The sine of any middle p)art is equal to the product 
of the cosines of the opposite parts. 

These rules are easily remembered by the expressions 
tan. ad. and cos. op. 

The correctness of these rules may be easily shown by 
taking in turn each of the five parts as middle 
part, and comparing the resulting equations with Co ^* 
the equations contained in Formulas [38] -[43], 
p. 144. 

For example, let Co. c be taken as middle part ; 
then Co. A and Co. B are the adjacent parts, and 
a and b the opposite parts, as is very plainly seen 
in Fig. 89. Then, by Napier's Rules, 

sin {Co. c) = tan (Co. A) tan (Co. B), 
or cos c = cot A cot B ; 

sin (Co. c) = cos a cos b, 
or cos c = cos a cos b. 

These results agree respectively with Formulas 
[43] and [38], p. 144. 




148 



SPHERICAL TRIGONOMETRY 



EXERCISE XXXIII 

1. Show that Napier's Rules lead to the equations contained 
in Formulas [39], [40], [41], and [42]. 

2. What will Napier's Rules become if we take as the five 
parts of the triangle the hypotenuse, the two oblique angles, 
and the complements of the two legs ? 



SECTION LII 

SOLUTION OF THE RIGHT SPHERICAL TRIANGLE 

By means of Formulas [38]-[43], p. 144, we can solve a 
right triangle in all possible cases. In every case two parts 
besides the right angle must be given. 





Fig. 90 

Case I 

Given the two legs a and b. 

From Formulas [38] and [42], p. 144, we obtain 
cos c = cos a cos b, 
tan A = tan a esc b, 
tan B = tan b esc a. 



THE RIGHT SPHERICAL TRIANGLE 



149 



For a check use cos c = cot A cot B, [43], p. 144. 

Example. Given a = 27° 28' 36", b = 51° 12' 8 
the triangle. 

log cos a = 9.94802 
log cos b = 9.79697 
log cos c 



solve 



9.74499 
= 56° 13' 41" 



log tan b = 10.09476 
log esc a = 0.33594 
log tan £ = 10.43070 
B = 69° 38' 54" 



log tan a = 9.71605 
log esc b = 0.10826 
log tan ,4 = 9.82431 

4 = 33° 42' 51" 



Check. 
log cot A = 10.17569 
log cot B = 9.56930 
log cos c = 9.74499 



Case II 

Given the hypotenuse c and the leg a. 
From Formulas [38], [39], and [40], p. 144, we obtain 
cos b = cos c sec a, 
sin A = sin a esc c, 
cos B = tan a cot c. 
For a check use cos B = cos b sin ^4, [41], p. 144. 
Although two angles in general correspond to sin A, one 
acute, the other obtuse, yet in this case it is easy to determine 
whether A is acute or obtuse since ,4 and a must be alike in 
kind. (See Example 2, Exercise XXXII, p. 146.) 



Case III 

Given the leg a and the opposite angle A. 

From Formulas [39], [42], and [41], we obtain 
sin c = sin a esc A, 
sin b = tan a cot A, 
sin B = sec a cos A. 



150 SPHERICAL TRIGONOMETRY 

Or, from [38] and [40], p. 144, we obtain 

cos b = cos c sec a, 
cos B — tan a cot c. 

For a check use sin b = sin c sin B, [39], p. 144. 

When c has been computed, b and B are determined by these 
values of their cosines ; but, since c must be found from its siue, c 
may have, in general, two values which are supplements of each 
other. This case, therefore, really admits of two solutions. 

In fact, if the sides b and c are extended until they meet in 
A' (Fig. 91), the two right triangles ABC and A'BC have the 
side a in common, and the angle A = A'. Also, A'C = 180° — b, 
A'B = 180° - c, and Z A 'BC = 180° - B. Hence, if ABC is 
one solution, A'BC is the other. 

Case IV 

Given the leg a and the adjacent angle B. 

From Formulas [40], [42], and [41], p. 144, we obtain 

tan c = tan a sec B, 

tan b = sin a tan B, 

cos A = cos a sin 5. 

For a check use cos A = tan & cot c, [40], p. 144. 

Case V 

Given the hypotenuse c erne? the angle A. 

From Formulas [39], [40], and [43], p. 144, we obtain 

sin a = sin c sin A, 

tan b = tan c cos A , 

cotB = cos c tan A 

Here a is determined by sin a, since <z and ^4 must be alike 
in kind. (See Example 2, Exercise XXXII, p. 146.) 
For a check use sin a === tan 6 cot 5, [42], p. 144. 



THE RIGHT SPHERICAL TRIANGLE 151 

Case VI 

Given the two angles A and B. 

From Formulas [43] and [41], p. 144, we obtain 

cos c = cot A cot B, 

cos a = cos A esc B, 

cos b — cos B esc A. 

For a check use cos c = cos a cos b, [38], p. 144. 

Note 1. In Case I (a and b given), if c is very near 0° or 180°, it may 
be found with greater accuracy by first computing B, and then computing 
c, as in Case IV. 

Note 2. In Case II (c and a given), if b is very near 0° or 180°, it 
may be computed more accurately by means of the derived formula 

tan 2 i b = tan £ (c - a) tan %(c + a). (Ex. 4, Sect. L) 

If A is so near 90° that it cannot be found accurately in the tables, it 
may be computed from the derived formula 

tan 2 (45° - £ A) = tan £ (c - a) cot \ (c + a). (Ex. 5, Sect. L) 

If B cannot be found accurately, we may use the formula 

tan 2 £J3 = sin(c — a) esc (c + a). (Ex. 6, Sect. L) 

Note 3. In Case III (a and A given), when the formulas do not give 
accurate results, we may employ the derived formulas 

tan 2 (45° - ic) = tan $(A-a) cot | (A + a), (Ex. 9, Sect. L) 

tan 2 (45° - £6) = sin (A - a) esc (A + «), (Ex. 10, Sect. L) 

tan 2 (45° - i B) = tan i (4 - a) tan -J- (4 + a) (Ex. 11, Sect. L) 

Note 4. In Case IV (a and B given), if A is near 0° or 180°, it may 
be more accurately found by first computing b and then finding A. 

Note 5. In Case V (c and A given), if a is near 90°, it may be found 
by first computing 6, and then computing a by Formula [42], p. 144. 

Note 6. In Case VI (A and B given), for unfavorable values of the 
sides greater accuracy may be obtained by means of the derived formulas 

tan 2 £c = - cos (A + B) sec (A - 5), (Ex. 7, Sect. L) 

tan 2 £a = tan [| (A + B) - 45°] tan [45° + $(A- £)], (Ex. 8, Sect. L) 
tan 2 1 b = tan [£ (A + B) - 45°] tan [45° -$(A- B)]. 



152 



SPHERICAL TRIGONOMETRY 



Note 7. In Cases I, IV, and V, the solution is always possible. 
In the other Cases, in order that the solution may be possible, it is 
necessary and sufficient that in Case II sin a < sin c ; in Case III, that 
a and A be alike in kind, and sin A > sin a ; in Case VI, that A + B + C 
> 180°, and the difference between A and B < 90°. 

Note 8. It is easy to trace analogies between the formulas for solving 
right spherical triangles and those for solving right plane triangles. The 
former become identical with the latter if we suppose the radius of 
the sphere to be infinite in length. Then the cosines of the sides 
become each equal to 1, and the ratios of the sines of the sides and 
of the tangents of the sides must be taken as equal to the ratios of the 
sides themselves. 

Note 9. In solving spherical triangles, the algebraic sign of the 
functions must receive careful attention. Write the sign of each func- 
tion just above the function. Then the sign of the function in the first 
member of the equation is + or — according as the law of signs makes 
the second member of the equation positive or negative. (See Example 1, 
p. 175.) 

If the function is a cosine, tangent, or cotangent, the + sign shows 
the angle < 90° ; the — sign shows the angle > 90°, and the supplement 
of the angle obtained from the table must be taken. 

If the function is a sine, since the sine of an angle and of its suppte- 
ment are the same, the acute angle obtained from the table and its 
supplement must be considered as solutions, unless there are other 
conditions that remove the ambiguity. For conditions 
that remove the ambiguity, see Examples 1 and 2 in 
Exercise XXXII, p. 146. 

It is always easy to find the required formula 
by means of Napier's Rules. In applying these 
rules we must choose for the middle part that one 
of the three parts which will make the two given 
parts either adjacent parts or opposite parts. 

Example : Given a and B ; solve the triangle. 

To find b, take a as the middle part ; then b and 
Co. B are the adjacent parts ; and, by Rule I, 
sin a = tan b cot B. 

Whence, tan b = sin a tan B. 



Co.B 




THE RIGHT SPHERICAL TRIANGLE 



153 



To find c, take Co. B as the middle part ; then a and Co. c 
are the adjacent parts ; and, by Rule I, 
cos B — tan a cot c. 



Whence, 



tan c = tan a sec B. 



To find A, take Co. A as the middle part; then a and Co. B 
are the opposite parts ; and, by Rule II, 
cos A = cos a sin B. 



EXERCISE XXXIV 



Solve the following right triangles, taking for the given 
parts in each case those printed in columns I and II : 





I 


II 


III 


IV 


V 




a 


b 


c 


A 


B 


1 


36° 27' 


43° 32' 31" 


54° 20' 


4G° 59' 43" 


57° 59' 19" 


2 


86° 40' 


32° 40' 


87° 11' 40" 


88° 11' 58" 


32° 42' 39" 


3 


50° 


30° 54' 49" 


59° 4' 20" 


G3° 15' 13" 


44° 20' 22" 


4 


120° 10' 


150° 59' 44" 


03° 55' 43" 


105° 44' 21" 


147° 19' 47" 




c 


a 


b 


A 


B 


5 


55° 9' 32" 


22° 15' 7" 


51° 53' 


27° 28' 38" 


73° 27' 11" 


6 


23° 49' 51" 


14° 16' 35" 


19° 17' 


37° 36' 49" 


54° 49' 23" 


7 


44° 33' 17" 


32° 9' 17" 


32° 41' 


49° 20' 16" 


50° 19' 10" 


8 


97° 13' 4" 


132° 14' 12" 


79° 13' 38" 


131° 43' 50" 


81° 58' 53" 




a 


A 


c 


b 


B 


9 


77 D 21'50" 


83° 56' 40" 


78°53'20'n 
101° 6' 40"} 


28°14'31" s l 
151° 45' 29"| 


28° 49' 57"! 
151° 10' 3") 


10 


77° 21' 50" 


40° 40' 40" 


impossible 







Note. The values in the last three columns of Example 9 cannot be 
combined promiscuously with those given in columns I and II. 

If a < 90°, with the value of b > 90° must be taken B > 90° and c > 90° ; 
while with the value of 6 < 90° must be taken, for the same reason, 
B < 90° and c < 90°. (See Examples 1 and 2, Exercise XXXII, p. 146.) 



154 



SPHERICAL TRIGONOMETRY 





I 


II 


III 


IV 


V 




a 


B 


c 


6 


A 


11 


92° 47' 32" 


50° 2' 1" 


91° 47' 40" 


49° 59' 58" 


92° 8' 23" 


12 


2° 0'55" 


12° 40' 


2° 3' 56" 


0° 27' 10" 


77° 20' 28" 


13 


20° 20' 20" 


38° 10' 10" 


25° 14' 38" 


15° 16' 50" 


54° 35' 17" 


14 


54° 30' 


35° 30' 


59° 51' 21" 


30° 8' 39" 


70° 17' 35" 




c 


A 


a 


b 


B 


15 


69° 25' 11" 


54°54 , 42" 


50° 


56° 50' 49" 


63° 25' 4" 


16 


112° 48' 


56° 11' 56" 


50° 


127° 4' 30" 


120° 3' 50" 


17 


46° 40' 12" 


370 40/ 9 // 


26° 27' 24" 


39° 57' 42" 


62° 0' 4" 


18 


118° 40' 1" 


128° 0' 4" 


136° 15' 32" 


48° 23' 38" 


58° 27' 4" 




A 


i? 


a 


b 


c 


19 


63° 15' 12" 


135° 33' 39" 


50° 0' 4" 


143° 5' 12" 


120° 55' 34" 


20 


116°43 , 12 // 


116° 31' 25" 


120° 10' 3" 


119° 59' 46" 


75° 26' 58" 


21 


46° 59' 42" 


57° 59' 17" 


36° 27' 


43° 32' 30" 


54° 20' 


22 


90° 


88° 24' 35" 


90° 


88° 24' 35" 


90° 



23. Define a quadrantal triangle, and show how its solution 
may be reduced to that of the right triangle. 

24. Solve the quadrantal triangle the sides of which are 

a = 174° 12' 49", b = 94° 8' 20", c = 90°. 

25. Solve the quadrantal triangle in which 

c = 90°, A = 110° 47' 50", B = 135° 35' 34.5". 

26. Given in a spherical triangle A, C, and c each equal to 
90° ; solve the triangle. 

27. Given A = 60°, C = 90°, and c = 90° ; solve the triangle. 

28. In a right spherical triangle, given A = 42° 24' 9", 
B = 9° 4' 11" ; solve the triangle. 

29. In a right spherical triangle, given a = 119° 11', 
B = 126° 54' j solve the triangle. 



THE RIGHT SPHERICAL TRIANGLE 155 

30. In a right spherical triangle, given c = 50°, b = 44°18'39"; 
solve the triangle. 

31. In a right spherical triangle, given A = 156° 20' 30", 
a = 65° 15' 45" ; solve the triangle. 

32. In a right spherical triangle, given A = 74° 12' 31", 

c = 64° 28' 47"; solve the triangle. 

33. In a right spherical triangle, given a = 112° 42' 38", 
B = 44° 28' 44" ; solve the triangle. 

34. In a right spherical triangle, given b = 48° 12' 48", 
A = 108° 14' 14" ; solve the triangle. 

35. In a right spherical triangle, given A = 122° 58' 47", 
B = 104° 17' 55" ; solve the triangle. 

36. If the legs a and b of a right spherical triangle are 
equal, show that cos a = cot A = Vcos c. 

37. In a right spherical triangle show that 

cos 2 A sin 2 c = sin (c + a) sin (c — a). 

38. In a right spherical triangle show that 

tan a cos c = sin 5 cot £. 

39. In a right spherical triangle show that 

sin 2 A = cos 2 B + sin 2 a sin 2 B. 

40. In a right spherical triangle show that 

sin (b + c) = 2 cos 2 £ .4 cos b sin c. 

41. In a right spherical triangle show that 

sin (c — b) = 2 sin 2 -| ^1 cos b sin c. 

42. If in a right spherical triangle, p denotes the arc of the 
great circle passing through the vertex of the right angle and 
perpendicular to the hypotenuse, m and n the segments of the 
hypotenuse made by this arc adjacent to the legs a and b, show 
that (i) tan 2 a = tan c tan ??i, (ii) sin 2 p = tan m tan n. 



156 



SPHERICAL TRIGONOMETRY 



SECTION LIII 

SOLUTION OF THE ISOSCELES SPHERICAL TRIANGLE 

An arc of a great circle, passed through the vertex of an 
isosceles spherical triangle and the middle point of the base 
(Fig. 94), divides the triangle into two equivalent right spher- 
ical triangles. Hence, the solution of an isosceles spherical 
triangle may be reduced to that of a right spherical triangle. 





Fig. 93 



Fig. 94 



Likewise the solution of a regular spherical polygon may be 
reduced to that of a right spherical triangle. Arcs of great 





Fig. 95 



Fig. 96 



circles, passed through the centre of the polygon and the ver- 
tices (Fig. 96), divide the polygon into equal isosceles triangles ; 



THE RIGHT SPHERICAL TRIANGLE 157 

and each one of these equal isosceles trian- 
gles may be divided into two equivalent 
right triangles. 

A regular spherical polygon (Fig. 97) is 
the polygon formed by the intersections of 
the spherical surface with the faces of a 
regular pyramid whose vertex is at the 
centre of the sphere. Fig. 97 

EXERCISE XXXV 

1. In an isosceles spherical triangle, given the base b and 
the side a ; find A the angle at the base, B the angle at the 
vertex, and h the altitude. 

2. In an equilateral spherical triangle, given the side a; 
find the angle A. 

3. Given the side a of a regular spherical polygon of n sides ; 
find the angle .1 of the polygon, the distance R from the centre 
of the polygon to one of the vertices, and the distance r from 
the centre to the middle point of one of the sides. 



Tetrahedron Hexahedron Octahedron Dodecahedron Icosahedron 

Fig. 98 

4. Compute the dihedral angles made by the faces of the 
five regular polyhedrons (Fig. 98). 

5. A spherical square is a regular spherical quadrilateral. 
Find the angle .4 of the square, having given the side a. 



CHAPTER VIII 
THE OBLIQUE SPHERICAL TRIANGLE 

SECTION LIV 



FUNDAMENTAL FORMULAS 

Let ABC (Fig. 99) be an oblique spherical triangle, a, b, c 
its three sides, A, B, C the angles opposite a, b, c, respectively. 

Through C draw CB, an arc of a 
great circle, perpendicular to the 
side AB, meeting AB at B. For 
brevity let 

CB=p, AD = n, BD = m, 
Z.ACB = x, ABCB = y. 

1. By [39], p. 144, in the right 
triangles BBC and ABC, 




and 



sin p = sin a sin B, 
sin p = sin b sin A. 



Therefore, 
similarly, 
and 



sin a sin B = sin b sin A 
sin a sin C = sin c sin A 
sin b sin C = sin c sin B 



[44] 



These equations may also be written in the form of 
proportions : 

sin a : sin b : sin c = sin A : sin B : sin C. 

That is, the sines of the sides of a spherical triangle are 
proportional to the sines of the opposite angles. 

158 



THE OBLIQUE SPHERICAL TRIANGLE 159 

In Fig. 99 the arc CD cuts the side AB within the triangle. 
If CD falls without the triangle, for instance to the right of 
CB, sin (180° — B) would then be employed instead of sin B. 
But sin (180° - B) = sin B (Sect. XXV, p. 48). Hence, the 
Formulas [44] hold true in all cases. 

2. In the right triangle BDC, by [38], p. 144, 

cos a = cosp cos m 

= cos^> cos(c — n) 
by [9], p. 56, = cos p cos c cos n + cos/? sin c sin n. 

Now, in the right triangle ADC, by [38], p. 144, 

cos p cos n = cos b. 
Whence, cos p = cos b sec n, 

and cos p sin n = cos b tan n. 

By [40], p. 144, tan n = tan b cos A. 

.'. cosp sin n = cos 6 tan b cos *4 

— sin b cos J . 

Substituting these values of cos^>cos?i and cos p sin n in 
the value of cos a, we obtain 

cos a = cos b cos c + sin b sinccos A "| 
and similarly, cos b = cos a cos c + sin a sin c cos B V . [45] 

cos c = cos a cos b + sin a sin b cos C J 

3. In the right triangle ADC, by [41], p. 144, 

cos J. = cos pj sin x 

— cos p sin(C — y) 

by [8], p. 56, = COSJ9 sin C cos y — cosp cos C sin y. 
Now, by [41], p. 144, 

cos p sin y = cos i?. 

Therefore, cos p = cos B esc y, 

and cos p cos y — cos B cot ?/ 

by [43], p. 144, = cos B tan B cos a 

= sin B cos a. 



160 SPHERICAL TRIGONOMETRY 

Substituting these values of cos p sin?/ and cos p cosy in 
the value of cos A , we obtain 

cos A = — cos B cos C + sin B sin C cos a "| 
and similarly, cos B = — cos A cos C + sin A sin C cos b ± • [4=6] 

cos C = — cos A cos B -f- sin A sin B cos c J 

Formulas [45] and [46] are also universally true ; for the 
same equations are obtained when the arc CD cuts the side 
AB without the triangle. 

EXERCISE XXXVI 

1. What do Formulas [44] become if A = 90° ? if B = 90°? 
if C = 90°? if <* = 90°? if A =B = 90°? iia = b = 90°? 

2. What do Formulas [45] become if A = 90° ? if B = 90° ? 
if C = 90° ? if A = B = C = 90 ? 

3. What does the first of Formulas [45] become if A =0°? 
if ^=90°? if ,4=180°? 

4. From Formulas [45] deduce Formulas [46], by means of 
the relations between polar triangles (Theorem 4, p. 141). 



SECTION LV 

FORMULAS FOR THE HALF ANGLES AND SIDES 

From the first equation of [45], p. 159, 

cos a — cos b cos c 



cos A = 
Therefore, 1 — cos A = 



sin b sin c 
sin b sin c + cos b cos c — cos a 
sin b sin c 



u rri-i K « cos (b — c)— cos a 

by r 9 l> P- 56, = ^ — r^ 

J L J r sin ft sin c 

by [23], p. 59, = -28i"*(» + ?-«)BinK»-«-« 3 . 

J . L J r . sin b sin c 



Also, 1 + cos A 
by [5], p. 54, 



THE OBLIQUE SPHERICAL TRIANGLE 161 

sin b sin c — cos b cos c -j- cos a 



sin 6 sin c 

_ cos a — cos (b + c) 
sin & sin c 



by [23], p. 59, 

Since by [16], p. 58 



— 2 sin i (a 4- b + c) sin \(a — b — c) 
sin & sin c 



1 — cos .1 = 2sin 2 i J, 
sin 2 ^A = sin £ (a -f b — c) sin \ (a — b + c) esc & esc c ; 
and since by [17], p. 58, 

1 + cos A = 2 cos 2 J A, 
cos 2 J A = sin J (a 4- # 4- c) sin £ (# 4- c — a) csc ^ csc c - 
Now let }(«4ft4c) = s. 

Then, %(b + c — a) = s — a t 

i(a — b + c) = s — b, 



and 



%(a + b -c) =s 



Then, by substitution and extraction of the square root, 



sin £ A = Vsin (s — b) sin (s — c) csc b csc c 
cos ^ A = Vsin s sin (s — a) csc b csc c 



[47] 



tan £ A = Vcsc s csc (s — a) sin (s — b) sin (s — c) 
In like manner, it may be shown that 



in J B = Vsin (s — a) sin (s — c) csc a csc c 



sin 



cos ^ B = Vsin s sin (s — b) csc a csc c 

tan -J- B = Vcsc s csc (s — b) sin (s — a) sin (s — c) 

sin i C = Vsin (s — a) sin (s — b) csc a csc b 

cos |C = Vsin s sin (s — c) csc a csc b 

tan J C = Vcsc s csc (s — c) sin (s — a) sin (s — b) j 



)•■ [ 4 7J 



162 



SPHERICAL TRIGONOMETRY 



Again, from the first equation of [46], p. 160, 
cos B cos C + cos .4 



cos a — 
Therefore, 

1 — cos a = 



sin B sin C 
sin B sin'C — cos B cos C — cos A 



by [5], p. 54, 

by [22], p. 59, = 

and 1 4- cos a = 



sin B sin C 

_ — cos (B + C) — cos J. 
sin j3 sin C 

- 2 cos fr (£ + C + ^) cos i (B 4- C - ^4) , 

sin B sin C 
sin B sin C + cos B cos C 4- cos A 



by [9], p. 56, = 



sin B sin C 
cos (B — C) 4- cos ^4 



sin 5 sin C 

by [22], p. 59, = 2°o»K*-C;M)<*>b*(B-C-»). 

J L J ' ^ ' sin 5 sin C 

Since by [16], p. 58, 

1 — cos a = 2 sin 2 £ a, 

sin 2 ia = — cos ^(5 4- C 4- -4) cos J (5 + C — ^cscJBcsc C; 

and since by [17], p. 58, 

1 4- cos a = 2 cos 2 ^ a, 

cos 2 i a = cos ^ (1* — C + ^4) cos ^ (B — C — A) esc 5 esc C. 

Now let J(^. 4- B 4- C)= 5. 



Then, 



5 (5+C-i)=S-i, 



K^l -5 + C)=S-B, 
and £(4 + 5 - C) = 5 - C. 

Then, by substitution and extraction of the square root, 

sin \ a = V — cos S cos (S — A) esc B esc C 



cos \ a = Vcos (S — B) cos (S — C) esc B esc C 



tan \ a = V— cos S cos (S — A) sec (S — B) sec (S — C) ^ 



• [48] 



THE OBLIQUE SPHERICAL TRIANGLE 



163 



And, in like manner, 



sin \ b = V — cos S cos (S — B) esc A esc C 
cos \ b = Vcos (S — A) cos (S — C) esc A esc C 



tan \ b = V— cos S cos (S — B) sec (S — A) sec (S — C) 
sin \ c = V — cos S cos (S — C) esc A esc B 



cos \ c = Vcos (S — A) cos (S — B) esc A esc B 

tan \ c = V— cos S cos (S — C) sec (S — A) sec (S — B) ^ 



• [48] 



SECTION LVI 



GAUSS'S EQUATIONS AND NAPIER'S ANALOGIES 

By [5], p. 54, 

cos \{A -f- B) = cos J A cos \ B — sin £ A sin $ B ; 

or, by substituting for cos^A, cos^-i?, sin ^ A, sin-^-i?, their 
values given in Formulas [47], p. 161, and reducing, 



w= 



/sin s sin (s — a) /sin s sin (s — b) 

cos i(A+B) = \ — A L X V : ^ } - 

% sm b sin c * sin a sin c 



/ sin (s — fr) sin (s — c) /sin (5 — a) sin (5 — c) 

V siri 7) sin n * sin /? sin n 



sm a sm c 



_ sin s / sin (s — a) sin (g — 6) _ sin (s — c) /sin (s — a) sin (s — b) 
sin c * sin a sin & sin c * sin a sin b 

_ sin 5 — sin (s — e) /sin (s — a) sin (s — 6) 
sin c " * sin a sin & 

By [21], p. 59, 

sin s — sin (s — c) = 2 cos J (s + s — c) sin £ (5 — s -f- c) 
= 2 cos (s — i- c) sin ^ c. 
By [12], p. 58, sin c = 2 sin ^ c cos £ c. 



164 



SPHERICAL TRIGONOMETRY 



Again, by [47], p. 161, 



4 



sin (s — a) sin (s 
sin a sin b 



*) 



= sin i C. 



Substituting in the value of cos \ (A -f- B), we have 



cos ±(A +B) 



_ 2 sin i c cos (s — i- c) 



sm 



sin ^ C. 



2 sin | c cos ^ c 
_ cos (s — -J- c) 
cos ^c 

.*. cos ^ (-4 + ■#) cos \c — cos (s — \ c) sin i C. 
Since s — ^c = |(a + 5), 

cos j- (J. + -B) cos |- c = cos |- (a + b) sin ^ C. 
By proceeding in like manner with the values of 

sin i (A + B), cos k(A — B), and sin £ (A — B), 

three analogous equations are obtained. 
The four equations, 

cos \ (A -+- B) cos \ c = cos ^ (a + b) sin ^ C " 
sini(A + B)cos|-c = cos J(a-b)cos|C j™ 

cos ^ (A — B) sin \ c = sin \ (a + b) sin \ C 
sin ^ (A — B) sin ^ c = sin £ (a — b) cos i C 

are called Gauss's Equations. 

By dividing the second of Gauss's Equations by the first, 
the fourth by the third, the third by the first, and the fourth 
by the second, we obtain 



tan.-J-(A.+ B) = 



«s s i n \ ( a 
tani(A-B)=- " 



cos -J- (a — b) 
cos i (a + b) 



tani(a + b) = 

tan ^ (a — b) = 



sin i (a + b) 
cos£(A — B) 
cos|(A + B) 
sin^(A-B) 
sin£(A + B) 



cot^C 


cot|C 


tan \ c 


tan^c 



[50] 



THE OBLIQUE SPHERICAL TRIANGLE 165 

There will be other forms in each case, according as other 
elements of the triangle are used. 

These equations are called Napier's Analogies. 

In the first equation the factors cos % {a — b) and cot £ C are 
always positive ; therefore, tan % (A + B) and cos % (a + b) 
must always have like signs. 

Hence, if a + b< 180°, cos £ (a + 6) > 0, and tan \ (A +B)>0. 
Hence, A + B < 180°. 

If a + b> 180°, then A + B > 180°. 

If a + b = 180°, cos i(a + b) = 0, and tan b(A + B) = oo. 
Hence, ±(.4 + J5) = 90°, and A + 5 = 180°. 

Conversely, it may be shown from the third equation, that 
a + b is less than, greater than, or equal to 180°, according as 
A + B is less than, greater than, or equal to 180°. 



SECTION LVII 
Case I 

Given two sides a and b and the included angle C. 

The angles A and B may be found by the first two of Napier's 
Analogies : 

After .4 and 5 have been found, the side c may be found by 
[44], p. 158, or by [50], p. 164 ; but it is better to use for this 
purpose Gauss's Equations, because they involve the functions 
of the same angles that occur in working Napier's Analogies. 
Any one of the equations may be used ; for example, 



166 



SPHERICAL TRIGONOMETRY 



cos \(a + b) . 

cos -i- c = f-y- A ~ sm 

2 cos i (A + B) 



ic. 



Example. Given a = 73° 58' 54", b = 38° 45', C 
solve the triangle. 

a = 73° 68' 54", 
6 = 38° 45' 0", 
C = 46° 33' 41". 



46°33'41"; 



log cos i (a- 6) = 9.97914 

log sec £ (a + 6) = 0.25658 

log cot £ C = 0.36626 

log tan i (A + B) = 0.60198 

log sec %{A + B) = 0.61515 

log cos £(a + b) = 9.74342 

log sin £ C = 9.59685 

log cos £c =9.95542 

£c =25°31 / 12 / 



Hence, £(a - 6) = 17° 36' 57", 
i(a + &) = 56°21' 57", 
i C = 23° 16" 50.5". 

logsini(a-6) = 9.48092 

log esc £ (a +6) = 0.07956 

log cot i C = 0.36626 

log tan i(A- B) = 9.92674 

±(A + B)= 75° 57' 40.8' 
i(A-B)= 40° 11' 25.4' 
(A = 116° 9' 6" 
*{ B = 35° 46' 15" 
L c = 51° 2' 24" 



To test the accuracy of the work we may use the Rule of 
Sine Proportion given in Sect. LIV, p. 158. 

If c only is desired, it may be found from [45], p. 159, 
without previously computing A and B. But the Formulas 
[45] are not adapted to logarithmic work. Instead of chang- 
ing them to forms suitable for logarithms, we may use the 

following method, which leads to 
the same results and has the ad- 
vantage that, in applying it, noth- 
ing has to be remembered except 
Napier's Rules : 

Through B (Fig. 100) draw an 
arc of a great circle perpendicu- 
lar to A C, cutting A C at D. Let 
BD=p, CD = m, AD = n. 
By Rule I, 

cos C = tan m cot a. 
Whence, tan m = tan a cos C. 




THE OBLIQUE SPHERICAL TRIANGLE 167 

By Rule II, 

cos a = cos m cos p ; whence, cos p = cos a sec m. 

cos c = cos n cos p ; whence, cos p = cos c sec w. 

Therefore, cos c sec w = cos a sec w. 

Since n = b — m, 

cos c = cos a sec m cos (& — m). 

Now c may be computed from the two equations 

tan m = tan a cos C ; 

cos c = cos a sec m cos (b — tti). 

Note. If BD falls without the triangle, for instance to the right of 
BC, then n = b + m. 

.• . cos c = cos a sec m cos (6 + m). 

Example. Given a = 97° 30', & = 55° 12', C = 39° 58'; findc. 



log tan a = 0.88057 (n) 
log cos C = 9.88447 
log tan m = 0. 76504 (n) 
m = 99° 44' 49' 
b - m = - 44° 32' 49' 



log cos a =9. 11570 (n) 
log seem = 0.77135 (n) 
cos (b - m) = 9.85289 
log cose = 9.73994 
c = 56° 40' 9" 



EXERCISE XXXVH 

1. What are the formulas for computing a when b, c, and A 
are given ; and for computing b when a, c, and B are given ? 

2. Given a = 88° 12' 20", J = 124° 7' 17", C = 50° 2' 1"; 

find A = 63° 15' 11", B = 132° 17' 58", c = 59° 4' 17". 

3. Given a = 120° 55' 35", b = 88° 12' 20", C = 47° 42' 1"; 

find A = 129° 58' 2", B = 63° 15' 8", c = 55° 52' 40". 

4. Given b = 63° 15' 12", c = 47° 42' 1", A = 59° 4' 25"; 

find B = 88° 12' 24", C = 55° 52' 42", a = 50° 1' 40". 

5. Given 6 -69° 25' 11", c = 109° 46' 19", ^ = 54° 54' 42"; 

find B = 56° 11' 57", C = 123° 21' 12", a = 67° 11' 47". 



168 



SPHERICAL TRIGONOMETRY 



SECTION LVIII 

Case II 

Given two angles A and B, and the included side c. 

The sides a and b may be found by the third and fourth of 
Napier's Analogies : 

, , , ,. cos \ (A — B) , 
ta a i(a + J) = coBt J il+jB jtani«i 

The angle C may be found by [44], p. 158, by Napier's 
second Analogy, [50], p. 164, or by one of Gauss's Equations, 
[49], p. 164. Thus, Gauss's second equation gives 

, „ sin-i-M + B). 

COS f C = ^ 7^ COS -i- C. 

2 cos £ (a — b) 2 



Example. Given.4 = 107°47'7",£: 
solve the triangle. 



:38°58'27",c = 51°41'14"; 



.4 = 107° 47' V 
B= 38° 58' 27' 
c= 51° 41' 14' 

log cos i(A-B) = 9.91648 

log sec ±(A + B) = 0.54359 

log tan £ c = 9.68517 

log tan i (a + 6) = 0.14524 

log sin i(^4 + 5) = 9.98146 

log sec $(a -b) = 0.01703 

log cos | c = 9.95423 

log cos $C = 9.95272 

A C = 26° 15' 10" 



Hence, £ (J. - 5) = 34° 24' 20", 

i{A + J5) = 73° 22' 47", 
i c = 25° 50' 37". 

log sin l(A-B) = 9.75208 

log esc i {A + B) = 0.01854 

log tan i c = 9.68517 

log tan £ (a - 6) = 9.45579 

i(a + 6) = 54°24'24.4 / 
i(a-6) = 15° 56' 25.5' 
a = 70° 20' 50" 
b =38° 27' 59" 
C = 52° 30' 20" 



THE OBLIQUE SPHERICAL TRIANGLE 



1G9 



If the angle C alone is wanted, we proceed as in Case I, p. 166, 
when the side c alone is desired. 

Let (Fig. 101) ZABD = x, Z.CBD = y, BD=p; then, 

Rule I, cos c = cot x cot A. 

Whence, cot x — tan A cos c. 
Rule II, cos A = cosp sin x. 
Whence, cos p = cos A esc x. 
Rule II, cos C = cosp sin y. 
Whence, cos/> = cos C esc y. 
.'. cos C = cos A esc # sin y 

= cos .4 esc x sin (Z? .— x) 
Now C niay be computed from the equations 

cot x = tan A cos c ; 

cos C = cos .1 esc # sin (B — xY 

Note. When BD falls to the right of BC, the last equation becomes 
cos C = cos A esc x sin (x — B). 




Example. Given A = 35° 46' 15", 5 = 115° 9' 



= 51°2': 



find C. 



log tan A =9.85760 
log cos c = 0.79856 
log cot x — 9.65616 

x = 65° 37' 35' 

.-. B - x = 49° 31' 32' 



log cos A= 9.90992 

log esc x =0.04056 

log sin (J5 - x) = 9.88121 

log cos C = 9.83069 

C = 47° 22' 42" 



EXERCISE XXXVm 



1. What are the formulas for computing A when B, C, and 
a are given; and for computing B when A, C, and b are given ? 

2. Given A = 26° oS' 46", B = 39° 45' 10", c = 154° 40' 48"; 

find a =37° 14' 10", b = 121° 28' 10", C= 161° 22' 11". 



170 SPHERICAL TRIGONOMETRY 

3. Given .4 = 128° 41' 49", £ = 107° 33' 20", c = 124°12'31"; 

find a = 125° 41' 43", b = 82° 47' 34", C = 127° 22'. 

4. Given B = 153° 17' 6", C = 78° 43' 36", a = 86° 15' 15"; 

find b = 152° 43' 51", c = 88° 12' 21", A = 78° 15' 48". 

5. Given A = 125° 41' 44", C = 82° 47' 35", ft = 52° 37' 57"; 

find a = 128° 41' 46", c = 107° 33' 20", 5 = 55° 47' 40". 



SECTION LIX 

Case III 

Given tivo sides a and b, and the angle A opposite a. 

The angle B is found from [44], p. 158 ; whence we have 

sin B = sin A sin b esc a. 

When B has been found, C and c may be found from the 
fourth and the second of Napier's Analogies : 

, . sinJ^A+i?) 

cotiC^^f^^tanH^-^ 
The third and first of Napier's Analogies may also be used. 

Note 1. Since B is determined from its sine, the problem in general 
has two solutions ; and, moreover, in case sin B > 1, the problem is impos- 
sible. By geometric construction it may be shown, as in the correspond- 
ing case in Plane Trigonometry, pp. 71-73, under what conditions the 
problem really has two solutions, one solution, or no solution. But in prac- 
tical applications a general knowledge of the shape of the triangle is known 
beforehand ; so that it is easy to see, without special investigation, which 
solution (if any) corresponds to the circumstances of the question. 

It can be shown that there are two solutions when A and a are alike 
in kind and sin &>sinei>sin^l sin 6 ; no solution when A and a are 
unlike in kind (including the case in which either A or a is 90°) and sin b 



THE OBLIQUE SPHERICAL TRIANGLE 



171 



is greater than sin a or equal to sin a, or when sin a is less than sin .4 sin b ; 
and one solution in every other case. 

Note 2. The side c or the angle C may be computed, without first 
finding B, by means of the formulas 

tan m = cos A tan 6, and cos (c — m) = cos a sec b cos m ; 
cot x = tan A cos 6, and cos (C — x) = cot a tan 6 cos x. 

These formulas may be obtained by resolving the triangle into right 
triangles, and then applying Napier's Rules ; m is equal to that part of 
the side c included between the vertex A and the foot of the perpendicu- 
lar from C, and x is equal to the corresponding portion of the angle C. 

Note 3. After the two values of B have been obtained, the number 
of solutions may be determined by Theorem 1, Sect. XLIX, p. 141. 

If log sin B is positive, there is no solution. 

Example. Given a = 57° 36', b = 31° 14', A = 104° 25' 30". 



In this case, A > 90°, 

and a + b < 180°. 

Therefore, A + B < 180°, 
and B< 90°. 

Hence, there is only one solution. 

a + b = 88° 50' 
a - b = 26° 22' 
A + B = 140° 55' 16" 
A - B = 67° 55' 44" 
log sin i(A + B) = 9.97424 
log esc i(A -B) = 0.25284 
log tan i(a - b) = 9.36966 
logtan£c = 9.59674 

i c = 21° 33' 37" 
c = 43° 7' 14" 



log sin A = 9.98609 
log sin b = 9.71477 
log esc a = 0.07349 
log sin E = 9.77435 

B = 36° 29' 46" 

£(a + &)=44°25' 
i(a-b) = 13°11' 
I {A + B) = 70° 27' 38' 
£(^L-£) = 33°57'52' 
log sin |(a+ b) = 9.84502 
log esc i(a - b) = 0.64194 
logtani(^L - B) = 9.82840 
logcot-£C = 0.31536 

i C = 25° 48' 58' 
C= 51° 37' 56' 



EXERCISE XXXIX 



Given a = 73° 49' 38", b = 120° 53' 35", A =88°52'42"; 

find 5 = 116° 42' 30", c = 120° 57' 27", C = 116° 47'. 
Given a = 150° 57' 5", b = 134° 15' 54", A =1U° 22 '42"; 

find ^ = 120°47'45", c, = 55° 42' 8", C\ = 97°42'55"; 
B 2 = 59° 12' 15", c 2 = 23° 57' 17", C 2 = 29°8'39". 



172 SPHERICAL TRIGONOMETRY 

3. Given a = 79° 0' 54", b = 82° 17' 4", A = 82° 9' 26" 

find B = 90°, c = 45° 12' 19", C = 45° 44' 5". 

4. Given a = 30° 52' 37", b = 31° 9' 16", A = 87° 34' 12" 
show that the triangle is impossible. 



SECTION LX 

Case IV 
Given two angles A and B, and the side a opposite A. 
The side b is found from [44], p. 158 ; whence, 

sin b = sin a sin 5 esc A. 
The values of c and C may then be found by means of the 
fourth and second of Napier's Analogies : 

, sin i (^ + £) , . . > ,. 

tail ^ = sinX(^-5) tan ^ a -^ ; 

2 sin J (a — 6) 2 v J 

Note 1. In this case the conditions for one solution, two solutions, or 
no solution can be deduced directly by the theory of polar triangles from 
the corresponding conditions of Case III, p. 170. There are two solutions 
when A and a are alike in kind and sin B > sin A > sin a sin B ; no solu- 
tion when A and a are unlike in kind (including the case in which either 
A or a is 90°) and sin B is greater than sin A or equal to sin A, or 
when sin A < sin a sin B ; and one solution in every other case. 

Note 2. By proceeding as indicated in Case III, Note 2, p. 171, formu- 
las for computing c or C, independent of the side &, may be found ; viz. , 
tan m — tan a cos B, and sin (c — m) = cot A tan 5 sin m ; 
cot cc = cos a tan 2?, and sin (C — x) = cos A sec 5 sin x. 
In these formulas ?n = BD, x = Z BCD, D being the foot of the per- 
pendicular from the vertex C. 

Note 3. As in Case III, p. 171, only those values of 6 can be retained 
which are greater than or less than a, according as B is greater than 
or less than A. If log sin 6 is positive, the triangle is impossible. 



THE OBLIQUE SPHERICAL TRIANGLE 173 

EXERCISE XL 

1. Given A =110° 10', B = 133° 18', a = 147° 5' 32"; 

find b = 155° 5' 18", c = 33° 1' 37", C = 70° 20' 40". 

2. Given ^=113° 39' 21", B = 123°40'18",a= 65° 39' 46"; 

find ft = 124° 7' 20", c = 159° 50' 15", C= 159° 43' 34". 

3. Given A = 100° 2' 11", 22 = 98° 30' 28", a = 95° 20' 39"; 

find b = 90°, c = 147°41'50",C= 148° 5' 40". 

4. Given 4 = 24° 33' 9", B = 38° 0' 12", a = 65° 20' 13" ; 
show that the triangle is impossible. 

SECTION LXI 
Case V 

Given the three sides a, b, and c. 

The angles are computed by means of Formulas [47], p. 161, 
and the corresponding formulas for the angles B and C. 

The formulas for the tangent are, in general, to be preferred. 
If we multiply the equation 



tan \ A = Vcsc s esc (s — a) sin (s — b) sin (s — c) 

U i-U 4.' 1 Sil1 ( S ~ a ) 

by the equation 1 = - — * r- > 

J ^ sin (s — a) 

and put tan r for Vcsc s sin (5 — a) sin (s — b) sin (5 — <•), at the 
same time making analogous changes in the equations for 
tan 1 B and tan ^ C, we obtain 

tan % A = tan r esc (5 — a), 
tan ^ B = tan r esc (s — b), 
tan i C = tan r esc (s — c), 

which are the most convenient formulas to employ when all 
three angles have to be computed. 



174 



SPHERICAL TRIGONOMETRY 



74 



Exa 


MP] 


jE 1. 


Given a 


°51' 


50 


'• find A. 








a = 


50° 54' 


32" 






b = 


37° 47' 


18" 






c = 


74° 51' 


50" 






2s = 


163° 33' 


40" 






s = 


81° 46' 


50" 




s - 


- a = 


30° 52' 


18" 




s - 


-b = 

- c = 


43° 59' 
6° 55' 


32" 



b = 37° 47' 18", c = 



log esc s = 0.00448 

log esc (s - a) = 0.28979 

log sin (s- 6) = 9.84171 

log sin (s - c) - 9.08072 

2 )19.21670 

logtani^4= 9.60835 

±A = 22° 5' 20" 
A = 44° 10' 40" 



Example 2. Given a = 124° 12' 31", b 

97° 12' 25" ; find ^ = 127° 22' 7", £ = 51 

26' 40". 

a - 124° 12' 31" 

b = 54° 18' 16" 
c = 97° 12' 25" 



= 54° 18' 16", c = 
18' 11", C = 72° 



2 s = 275° 43' 12' 

s = 137° 51' 36' 

log sin (s -a) = 9.37293 

log sin (s- 6) =9.99725 

logsin(s-c) =9.81390 

log esc s = 0.17331 

log tan 2 r = 9.35739 

log tan r = 9.67870 



s - a = 13° 39' 5" 
s - b = 83° 33' 20" 
s - c = 40° 39' 11" 
log tan £^4 = 0.30577 
log tan ££ =9.68145 
log tan i C = 9.86480 

iA= 63° 41' 3.8' 

iB = 25° 39' 5.6' 

iC= 36° 13' 20" 

^1 = 127° 22' 7" 

5= 51° 18' 11" 

C= 72° 26' 40" 



EXERCISE XLI 



Given a 

find .4 
Given a 

find J. 
3. Given a 

find ^ 
Given a- 

find ,4 



2. 



4. 



120°55'35", 6 = 59°4'25", c 
116°44'50", £ = 63°15'10", C 

50°12'4", b = 116°44'48", c 
59°4'28", £ = 94°23'12", C 
131°35'4", & = 108°30'14", c 
132°14'21", B = 110° 10' 40", C 



106 J 10'22"; 

91°7'22". 

129°11'42"; 

120°4'52". 

84°46'34"; 

99°42'24". 

20°16'38", & = 56°19'40", c = 66°20'44"; 
20°9'55", £ = 55° 52' 35", C = 1U°20'21". 



THE OBLIQUE SPHERICAL TRIANGLE 175 

SECTION LXII 
Case VI 

Given the three awjles A, B, and C. 

The sides are computed by means of Formulas [48], p. 162. 
The formulas for the tangents are, in general, to be preferred. 
If we multiply the equation 

tan i a = V— cos S cos (S — A) sec (S — B) sec (S — C) 

by the equation 1 = \- -f > 

J H sec(S-A) 



and put tan R for V— cos S sec (S — A) sec (S — B) sec (S — C), 

at the same time making analogous changes in the equations 
for tan J b and tan % c, we obtain 

tan ^ a = tan R cos (8 — A), 
tan \ b = tan R cos (S — 1?), 
tan \ c = tan R cos (5 — C), 

which are the most convenient formulas to use in case all 
three sides have to be computed. 

Example 1. Given A = 220°, B = 130°, C = 150°; find a. 

After we find the values of S, S — A, S — B, S — C, we write the 
formula for tan £ a with the algebraic sign written above each function 



as fol 


lows : 
tan \a 












= \ - cos S 


+ 
cos 


(S- 


- A)sec(S - B)sec(S - C). 






A = 220° 






log cos 5 = 9.53405 (n) 






B = 130° 






log cos (S -A) = 9.93753 






C = 150° 






log sec (5-5) = 0.30103 (w) 






2S = 500° 






log sec (S- C) =0.76033 (?i) 






S = 250° 






2)0.53294 




S - 


- A = 30° 






log tan £ a = 0.26647 




s 


- B = 120° 






ia= 61° 34' 6' 




s 


- C = 100° 






a = 123° 8' 12' 



176 



SPHERICAL TRIGONOMETRY 



Note. Here the effect, as regards algebraic sign, of three negative 
factors is canceled by the negative sign before the whole product. 

Example 2. Given A = 20° 9' 56", B = 55° 52' 32", C = 
114°20'14"; find a = 20°16'38", 6 = 56°19'41", c = 66°20'43". 

After we find the values of S, S — A, S — B, S — C, we write the 
formula for tan R with the algebraic sign written above each function 
as follows : 



tanE 



=v 



cos S sec (S - A) sec (S - B) sec (S - C). 



A = 20° 9' 56' 
B= 55°52 / 32 / 
C = 114° 20' 14' 



2 S = 190° 22' 42" 

log cos 5 = 8.95638 (n) 

logsec(S-^i) = 0.58768 

log sec (S - £) =0.11143 

log sec (S- C) = 0.02472 

log tan 2 E = 9.68021 

log tan R =9.84010 



S = 95° 11' 21' 
S-A= 75° 1'25' 
S - B= 39° 18' 49' 
S - C = - 19° 8' 53' 
log tan £ a = 9.25242 
log tan $ b =9.72867 
log tan £ c = 9.81538 

£a = 10° 8' 18.9" 
£6 = 28° 9' 50.3" 
$ c = 33° 10' 21.4" 



a = 20° 16' 38' 
b = 56° 19' 41' 
c = 66° 20' 43' 



EXERCISE XLII 



1. Given A = 130 c 



B = 110°, 



80 c 



find a = 139° 21' 22", b = 126° 57' 52", c = 56° 51' 48". 

2. Given A = 59° 55' 10", B = 85° 36' 50", C = 59° 55' 10" ; 

find a = 51° 17' 31", b = 64° 2' 47", c = 51° 17' 31". 

3. Given A = 102° 14' 12", B = 54° 32' 24", C = 89° 5' 46"; 

find a = 104° 25' 9", 6 = 53° 49' 25", c = 97°44'19". 

4. Given A = 4° 23' 35", B = 8° 28' 20", C = 172° 17' 56" ; 

find a = 31° 9' 13", b = 84° 18' 28", c = 115° 10'. 



THE OBLIQUE SPHERICAL TRIANGLE 177 

SECTION LXIII 
AREA OF A SPHERICAL TRIANGLE 

I. When the three angles A, B, C are given. 
Let R = the radius of the sphere, 

E = the spherical excess = A + B + C — 180°, 
F = the area of the triangle. 

Three planes passed through the centre of a sphere, each 
perpendicular to the other two planes, divide the surface of 
the sphere into eight tri-rectangular triangles (Geometry, § 802). 

It is convenient to divide each of these eight triangles into 
90 equal parts, and to call each of these equal parts a spherical 
degree. The surface of every sphere, therefore, contains 720 
spherical degrees. 

Now in spherical degrees, the AABC = E (Geometry, § 834), 
and the surface of the sphere is equal to 720 spherical degrees. 

Hence, A ABC : surface of the sphere = E : 720. 

Since the surface of a sphere = 4 ttR 2 (Geometry, § 824), 

A ABC : 4: wR* = E :720. 

ttR 2 F 
Whence, F=— — • T51] 

180 L J 

II. When the three sides a, b, c are given. 

A formula for computing the area is deduced as follows : 
From the first of Formulas [49], p. 164, 

cos £ (A + B) _ cos \ (a + b) 
sin i C cos i c 

Now, sin i C = cos (90° - i C). 

Therefore, m W\% = "* H \ +S > - 

cos (90 — i C) cos i c 



178 SPHERICAL TRIGONOMETRY 



Then, by division and composition, 

cos \ (A + B) — cos (90° — j C) _ cos j- (a + b) — cos j c 
cos b(A+B) + cos (90° — £ C) ~ cos£(a + #) + cos^c 

Dividing [23], p. 59, by [22], p. 59, 

cos A — cos i? 



(a) 
- tan b(A+B) tan J- (A - B). (b) 



cos A + cos B 

Substituting in (b) for A and B, f(A + B) and 90° - JC, 
respectively, we have 
cos $(A + B)- cos (90° -jC) 
cos i (^ + B) + cos (90° - 1 C) 

= -tani(l^+i£ + 90 o -iC)tani(i.4+i J B-90° + iC) 

= -tan±(4 +B- C + 180°) tan ±(4 -f £ + C - 180°). 

Now, E = A+B+C- 180°. 

. *.tani(^ +^- C + 180°) = tan£ (360°- 2C+.4+5+C-180 )' 

= tani(360°-2C + £) 

= tan[90°-i(2C-^)] 

= cot£(2C-S). 

Now, substituting EfoiA+B+C — 180° and cot i(2C-E) 
for tan i(A + 5 - C + 180°), we have 

cos i (4 + B) + cos (90° — £C) 4V J 4 w 

Again substituting, in equation (b), for ^4 and B the values 
£ (a + 5) and -J- c, and also substitute 5 f or J (a + b + c) and 
s — c for J (a + b — c), we have 



cos|( a + 5)-cos|c = _ tan ta 
cos b (a + #) + cos ^- c 



c). (d) 



Comparing (a), (c), and (d), we obtain 

cot i (2 C — E) tan £ £ = tan | s tan £ (s — c). (e) 



THE OBLIQUE SPHERICAL TRIANGLE 



179 



By beginning with the second equation of [49], p. 164, and 
treating it in the same way, we obtain as the result 

tan±(2C - £)tan \E = tan -J- (s - a) tan £ (s - 6). (f) 

By taking the product of (e) and (f) we obtain the elegant 
formula, known as l'Huilier's Formula, 

tan 2 £ E = tan £ s tan \ (s — a) tan \ (s — b) tan \ (s — c). [52] 

By means of [52] E may be computed from the three sides, 
and then the area of the triangle may be found by [51], p. 177. 

III. In all other cases, the area may be found by first solving 
the triangle so far as to obtain the angles or the sides, which- 
ever may be more convenient, and then applying [51] or [52]. 

Example 1. Given A = 102° 14' 12", B = 54° 32' 24", 
C = 89° 5' 46" ; find the area of the triangle. 



A = 102° 14' 12' 
B= 54° 32' 24' 
C= 89° 5' 46' 



245° 52' 22' 
E = 65° 52' 22' 
= 237 142" 
180° = 648000" 



log 



log R 2 = log HP- 
log E= 5.37501 
it 



648000 



4.68557 - 10 



F =0.06058 + log R 2 
F=1.1497E 2 



Hence, if we know the radius of the sphere, we can express 
the area of a spherical triangle in the ordinary units of area. 

Example 2. Given a = 133° 26' 19", b = 64° 50' 53", c = 
144° 13' 45"; find E = 200° 46' 46". 

logtan£s = 1.11669 

logtan£(s — a) = 9.53474 

log tan | (s - b) - 0. 12612 

log tan -fc (s - c) = 9.38083 

log tan 2 i^= 0.15838 

log tan IE = 0.07919 

±E= 50° 11' 41. 5" 
# = 200° 46' 46" 
* See Wentworth & Hill's Logarithmic and Trigonometric Tables, p. 20. 



a = 133° 26' 19" 
b = 64° 50' 53" 
c = 144°13 / 45" 
2 s = 342° 30' 57" 
s= 171° 15' 28.5' 
s -a= 37° 49' 9.5' 
s -b = 106° 24' 35. 5' 
s-c= 27° i'43.5' 



180 SPHERICAL TRIGONOMETRY 

EXERCISE XLIII 

1. Given A = 84° 20' 19", B = 27° 22' 40", C = 75° 33' ; find 
E = 26159", F = 0.12682 R\ 

2. Given a = 69° 15' 6", b = 120° 42' 47", c = 159° 18' 33"; 
find E = 216° 40' 18". 

3. Given a = 33° 1' 45", b = 155° 5' 18", C = 110° 10'; find 
£ = 133° 48' 53". 

4. Given c = 114° 27' 57", 4 = 78° 42' 33", B = 127° 13' 7"; 
find the area. 

5. Given a = 76° 14' 47", b = 82° 40' 15", A = 60° 22' 44"; 
find the area. 

6. Given A = 80° 12' 35", B = 77° 38' 22", a = 76° 42' 28"; 
find the area. 

7. Given b = 44° 27' 40", c = 15° 22' 44", A = 167° 42' 27"; 
find the area. 

8. Given b = 67° 15' 42", A = 84° 55' 8", C = 96° 18' 49" ; 
find the area. 

9. Given b = 72° 19' 38", c = 54° 58' 52", B = 77° 15' 14"; 
find the area. 

10. Given B = 127° 16' 4", C = 42° 34' 19", 6 = 54° 47' 55"; 
find the area. 

11. Given a = 128° 42' 56", b = 107° 13' 48", c = 88° 37' 51"; 
find the area. 

12. Given ,4 =127° 22' 28", £ = 131°45'27", C = 100°52'16"; 
find the area. 

13. Givena = 116°19'45",^=160°42'24", C = 171°27'15"; 
find the area. 

14. Find the area of a triangle on the earth's surface 
(regarded as spherical) if each side of the triangle is equal 
to 1°. (Radius of earth = 3958 miles.) 



CHAPTER IX 

APPLICATIONS OF SPHERICAL TRIGONOMETRY 

SECTION LXIV 

PROBLEM 

To reduce to the horizon an angle measured in space. 

Let (Fig. 102) be the position of the eye ; DOC = h, the 
angle measured in space ; OD' and OC the projections of the 
sides of the angle upon the horizontal plane ; DOD' = m and 




p|f^^P-':-gi.'g"J 



B :; -in 



Fig. 102 



COC = n, the angles of inclination of OD and OC, respec- 
tively, to the horizon. Required the angle D'OC = x made 
by the projections on the horizon. 

181 



182 



SPHERICAL TRIGONOMETRY 



The planes of the angles BOD' and COC intersect in the 
line OP, perpendicular to the horizontal plane (Wentworth's 
Geometry, § 556). 

From as a centre describe a sphere, and let its surface cut 
the edges of the trihedral angle O-DCP in M, N, and P. 

In the spherical triangle MNP, MN=h, MP = 90° - m, 
jVP = 90° — n are known ; and P = x is required. 

By [47], p. 161, 



cos hx 



: \ sin /'QO _ m \ sin f90° — n\ 



sin (90° -m) sin (90° -n) 
Putting % (h + m + n) = s, we obtain 



cos ix = y j ™ P>«° + (,-*)! sin (90° -s) 



cos m cos ?i 



= Vcos (s — A) cos s sec m sec ?i. 



SECTION LXV 
PROBLEM 



To find the distance between two places on the earth 1 s surface 
(regarded as spherical), given the latitudes of the places and the 
difference of their longitudes. 




Fig. 103 



Fig. 104 



APPLICATIONS 



183 



Let M and N (Fig. 104) be the places. Then the distance 
MN is an arc of the great circle passing through the places. 
Let P be the pole, ARB the equator. The arcs MR and NS 
are the latitudes of the places, and the arc RS, or the angle 
MPN, is the difference of their longitudes. Let MR = b, 
NS = a, RS — I; then in the spherical triangle MNP two 
sides, MP — 90° — b, NP = 90° — a, and the included angle 
MPN = I are given, and we have from Sect. LVII, p. 167, 

tan m — cot a cos /, 
cos MN = sin a sec m sin (b + m). 

From these equations first find m, then the arc MN, and 
then reduce MN to geographical miles, of which there are 60 
in each degree. 



SECTION LXVI 



THE CELESTIAL SPHERE 



The Celestial Sphere is an imaginary sphere of indefinite 
radius, upon the concave 
surface of which all the 
heavenly bodies appear to 
be situated. 

The Celestial Equator, or 
Equinoctial, AVBU (Fig. 
105), is the great circle in 
which the plane of the 
earth's equator intersects 
the surface of the celestial 
sphere. 

The Poles. P and P' 
(Fig. 105), of the celestial 
equator are the points in which the earth's axis produced cuts 
the surface of the celestial sphere. 




184 



SPHERICAL TRIGONOMETRY 



The Celestial Meridian, PBP' (Fig. 106), of an observer is the 
great circle in which the plane of his terrestrial meridian 
produced meets the surface of the celestial sphere. 

Hour Circles, as PMP' (Fig. 106), or Circles of Declination, are 
great circles passing through the poles, perpendicular to the 
celestial equator. 

The Horizon, NWSE (Fig. 106), of an observer is the great 
circle in which the plane tangent to the earth's surface, at the 
place where he is, meets the surface of the celestial sphere. 

The Zenith, Z (Fig. 106), of an observer is that pole of his 
horizon which is exactly above his head. 

Vertical Circles, as NPZS (Fig. 106), are great circles pass- 
ing through the zenith of an observer, perpendicular to his 
horizon. 

The vertical circle passing through the east and west points 
of the horizon is called the Prime Vertical; that passing 
through the north and south points coincides with the celes- 
tial meridian. 

Q 
Z 




Fig. 106 



The Ecliptic, A VB (Fig. 107), is a great circle of the celes- 
tial sphere, apparently traversed by the sun in one year from 
west to east, in consequence of the motion of the earth around 
the sun. 



APPLICATIONS 



185 



The Equinoxes are the points where the ecliptic cuts the 
celestial equator. They are distinguished as the Vernal 
equinox and the Autumnal equinox ; the sun in his annual 
journey passes through the former on March 21, and through 
the latter on September 21. The Vernal equinox is shown 
here at V (Fig. 108). 

Circles of Latitude, as QMT (Fig. 108), are great circles passing 
through the poles of the 
ecliptic, perpendicular to 
the plane of the ecliptic. 

The angle which the 
ecliptic makes with the 
celestial equator is called 
the obliquity of the ecliptic; 
it is equal to 23° 27', nearly, 
and is often denoted by the 
letter e. 

The earth's diurnal mo- 
tion causes all the heavenly 
bodies to appear to rotate 
from east to west at the 
uniform rate of 15° per hour. If in Fig. 106 we conceive the 
observer placed at the centre 0, and his zenith, horizon, and 
celestial meridian fixed in position, and all the heavenly 
bodies rotating from east to west around PP' as an axis at 
the rate of 15° per hour, we form a correct idea of the 
apparent diurnal motions of these bodies. When the sun or 
a star in its diurnal motion crosses the meridian, it is said to 
make a transit across the meridian ; when it passes across the 
part NWS of the horizon, it is said to set ; and when it passes 
across the part NES, it is said to rise (the effect of refraction 
being here neglected). Each star, as M, describes daily a 
small circle of the sphere parallel to the celestial equator, and 
called the diurnal circle of the star. The nearer the star is to 




186 



SPHERICAL TRIGONOMETRY 



the pole, the smaller is the diurnal circle ; and if there were 
stars at the poles P and P', they would have no diurnal 
motion. To an observer north of the equator the north pole P 
is elevated above the horizon (Fig. 108) ; to an observer south 
of the equator the south pole P' is the elevated pole. 



SECTION LXVII 
SPHERICAL CO-ORDINATES 

Several systems of fixing the position of a star on the sur- 
face of the celestial sphere at any instant are in use. In each 
system a great circle and its pole are taken as standards of 
reference, and the position of the star is determined by 
means of two quantities called its spherical co-ordinates. 

I. If the horizon and the zenith are chosen (Fig. 109), the 
co-ordinates of the star are called its altitude and its azimuth. 





Fig. 110 



The Altitude of a star is its angular distance, DM, measured 
on a vertical circle, above the horizon. The complement, MZ, 
of the altitude is called the Zenith Distance. 

The Azimuth of a star is the angle PZM at the zenith formed 
by the meridian of the observer and the vertical circle passing 
through the star, and is measured, therefore, by an arc of the 
horizon. It is usually reckoned from the north point of the 



APPLICATIONS 187 

horizon in north latitudes, and from the south point in south 
latitudes ; and east or west according as the star is east or 
west of the meridian. 

II. If the celestial equator and its pole are chosen (Fig. 110), 
then the position of the star may be fixed by means of its decli- 
nation and its hour angle. 

The Declination of a star is its angular distance, RM, from 
the celestial equator, measured on an hour circle. The angular 
distance, PM, of the star, measured on the hour circle, from the 
elevated pole, is called its Polar Distance. 

The declination of a star, like the latitude of a place on the 
earth's surface, may be either north or south ; but, in practical 
problems, while latitude is always to be considered positive, 
declination, if of a different name from the latitude, must be 
regarded as negative. 

If the declination is negative, the polar distance is equal 
numerically to 90° + the declination. 

The Hour Angle of a star is the angle MPQ at the pole formed 
by the meridian, APE. of the observer and the hour circle, PMR. 
passing through the star. On account of the diurnal rotation 
the hour angle is constantly changing at the rate of 15° per 
hour. Hour angles are reckoned from the celestial meridian, 
positive towards the west, and negative towards the east. 

III. The celestial equator and its pole being still retained, we 
may employ as the co-ordinates of the star its declination and 
its right ascension. 

The Right Ascension of a star is the arc VR of the celestial 
equator included between the Vernal equinox and the point 
where the hour circle of the star cuts the celestial equator. 
Eight ascension is reckoned from the Vernal equinox eastward 
from 0° to 360°. 

IV. The ecliptic, EVF, and its pole Q may be taken as the 
standards of reference. The co-ordinates of the star are then 
called its latitude and its longitude. 



188 



SPHERICAL TRIGONOMETRY 



The Latitude of a star is its angular distance, MT (Fig. 112), 
from the ecliptic measured on a circle of latitude. 

The Longitude of a star is the arc VT (Fig. 112) of the 
ecliptic included between the Vernal equinox and the point 
where the circle of latitude through the star cuts the ecliptic. 
The longitude of a star is always measured eastward from the 
Vernal equinox. 





Fig. 112 



For the star M (Figs. Ill and 112), let 
I = the latitude of the observer, 
h = DM = the altitude of the star (Fig. Ill), 
z — ZM = the zenith distance of the star, 
a = Z PZM = the azimuth of the star, 
t = Z ZPM = the hour angle of the star, 
d = RM = the declination of the star, 
p = PM = the polar distance of the star, 
r = VR = the right ascension of the star (Fig. 112), 

u = MT = the latitude of the star, 
v = VT = the longitude of the star, 

NZS = the celestial meridian (Fig. Ill), 
ARB — the celestial equator (Fig. 112), 
EVF = the ecliptic. 



APPLICATIONS 



189 



In many problems a simple way of representing the mag- 
nitudes involved is to project the sphere on the plane of the 
horizon, as shown in Figs. 114 and 115. 





Fig. 113 



Fig. 114 



NESW is the horizon, Z the zenith, 
NZS the celestial meridian, WZE the 
prime vertical, WBE an arc of the 
celestial equator, P the pole, M a star, 
DM its altitude, ZM its zenith dis- 
tance, Z PZM its azimuth, MR its 
declination, PM its polar distance, 
Z ZPM its hour angle. 

SECTION LXVITI 







E 




m 


p' 


-M- 


-Ms 


\ 


\ 




\\ 


W D' 





Fig. 115 



THE ASTRONOMICAL TRIANGLE 

The triangle ZPM (Fig. Ill) is often called the astronom- 
ical triangle, on account of its importance in problems in 
Nautical Astronomy. 

The side PZ is equal to the complement of the latitude of 
the observer. For (Fig. Ill) since is the centre of the 
sphere, the angle ZOB between the zenith of the observer 
and the celestial equator is obviously equal to his latitude, 



190 



SPHERICAL TRIGONOMETRY 




and the angle POZ is the complement of ZOB. Since the arc 
NP is the complement of PZ, the altitude of the elevated pole 
is equal to the latitude of the place of observation. 

The triangle ZPM then (however much it may vary in 
shape for different positions of the star M) always contains 
the following five magnitudes : 

PZ = the co-latitude of observer 

= 90° - I, 
ZM = the zenith distance of star 
= z, 
Z. PZM = the azimuth of star 
= a, 
PM = the polar distance of star 

= P> 

Z. ZPM = the hour angle of star 
= t. 

A very simple relation exists between the hour angle of the 
sun and the local (apparent) time of day. The hourly rate at 
which the sun appears to move from east to west is 15°, and it 
is apparent noon at any place when the sun is on the meridian 
of that place. Hence, it is evident that if the hour angle is 
0°, the time of day is noon ; if the hour angle is 15°, the 
time of day is 1 o'clock p.m. ; if the hour angle is 75°, the 
time of day is 5 o'clock p.m. ; if the hour angle is — 15°, 
the time of day is 11 o'clock a.m. ; if the hour angle is — 75°, 
the time of day is 7 o'clock a.m. ; and so on. 

In general, if t denotes the absolute value of the hour angle, 
when the sun is west of the meridian, 

the time of day is — p.m. ; 

when the sun is east of the meridian, 

the time of day is 12 — — a.m. 



APPLICATIONS 



191 



SECTION LXIX 



PROBLEM 




Given the latitude of the observer and the altitude and the 
azimuth of a star, to find its declination and its hour angle. 

In the triangle ZPM (Fig. 117), 
given 

PZ = 90° - I = the co-latitude, 
ZM = 90° - h = the co-altitude, 
Z PZM = a = the azimuth ; 

to find 

PM = 90° — d — the polar distance, 
Z ZPM — t = the hour angle. 

Draw MK _L to NZS. 
Let ZK = vi. 

Then, if a < 90°, PK = 90° - (I -f m) j 
and if a > 90°, PA' = 90° - (J - m). 

By Napier's Eules, 

cos a = ± tan m tan /?, 
sin d = cos PA cos J/A, 
sin /i = cos ?>* cos MK. 
From (1), tan m = ± cot /> cos a. 

From (3), cos MK = sin h sec m. 

From (2), sin rf = cos [90° - (l± m)] cos MA. 

. • . sin e? = sin (I ± m) cos MfiT. 
Substitute in (5) the value of cos MK. 
Then, sin d = sin (I ± m) sin /* sec m. 

In equations (A) and (B) the — sign is to be used if a > 90°. 
The hour angle may then be found by means of [44], p. 158, 
whence sin t = sin a cos h sec d. 



(i) 

(2) 
(3) 
(A) 
(-t) 

(5) 
(B) 



192 



SPHERICAL TRIGONOMETRY 



SECTION LXX 



PROBLEM 



To find the hour angle of a heavenly body when its declina- 
tion, its altitude, and the latitude of the place are known. 

7 ^ In the triangle ZPM (Fig. 118), 

given PZ = 90° - I, 

M\\ K PM = 90° - d = *> 



1 E_\ 


\fe \ 


^ ^ — „„ ^ — ±J , 
ZM = 90° - h ; 


y^- y 


D J / 


required 




Z.ZPM=t 


^v / s*p' 


If, in the first formula of [47], 


Z' 


p. 161, 


Fig. 118 


sin i A 




= Vsin (.9 — b) sin (5 — c) esc b esc c, 


we put A — t, a = 90° 


- h, b=p, c = 90° -I, 


and 


2s = a + b + c. 


Then, 


2s = 90°-h+p + 90°-l, 


or 


2 s = 180°- I +p-h-, 


whence, 


s = c)0 o -il + ip-ih, 


s 


-J = 90°-J(i+^ + A), 


s 


-o = \(l+p-K)\ 


and the formula becomes 


sin J t 




= ± Vsin[90 o -i(Z+j9 + A)]sin^(Z+^-/i)cscj9CSc(90°-Z) 



= ± Vcos \ (I -\-p + h) sin 1 (£ -f^? — A) cscp sec I, 

in which the — sign is to be taken when the body is east of 
the meridian. 

If the body is the sun, how can the local time be found when 
the hour angle has been computed (Sect. LXVIII, p. 189) ? 



APPLICATIONS 193 

SECTION LXXI 

PROBLEM 

To find the altitude and the azimuth of a celestial body when its 
declination, its hour angle, and the latitude of the place are known. 

In the triangle ZPM (Fig. 118), 
given PZ = 90° - I, 

PM = 90° — d=p, 

Z ZPM = t ; 
required ZM = 90° - h, 

Z ZPM = a. 
Draw MK± to NZS and let PK = m. 
Then, if a < 90°, ZK = 90° - (I + w); 
and if a > 90°, ZK = (I + »&) - 90°. 

By Napier's Rules, 

cos £ = tan ?/i tan d, (1) 

sin A = sin (I + >/?) cos MK, (2) 

sin (Z = cos MK cos ?», (3) 

cos (7 -J- m) = cot a tan MJT, (4) 

sin ??i = cot t tan J/ A'. (5) 

From (1), tan m = cot d cos f. (A) 

From (3), cos MK = sin 6? sec w. (6) 

Substitute in (2) the value of cos MK, 

sin h = sin (/ -h m) sin d sec »*. (B) 

From (5), tan MK = sin m tan £. 

Substitute in (4) the value of tan MK, 

tan a == sec (I + w) sin m tan £. (C) 

Equations (A), (B), and (C) are the equations required. 
In (C) a is E. or W. to agree with the hour angle. 



194 



SPHERICAL TRIGONOMETRY 



SECTION LXXII 




PROBLEM 

To find the latitude of the place when the altitude of a celes- 
tial body, its declination, and its hour angle are known. 

In the triangle ZPM (Fig. 119), 
given ZM = 90° - h, 

PM = 90° - d, 
Z ZPM = t ; 
required PZ = 90° - I. 

Draw MK _L to NZS. 
Let PK = m, ZK = n. 

Then, by Napier's Rules, 
cos t = tan m tan d, 
sin h = cos n cos MK, 
sin d = cos m cos MK ; 
tan??i= cot d cos t, 
cos n = cos m sin h esc ^. 
It is evident from the figure that 

I = 90° - ( m ± »), 

in which the sign + or the sign — is to be taken according as 
the celestial body and the elevated pole are on the same side 
of the prime vertical or on opposite sides. 

In fact, both the values of I shown above may be possible 
for the same altitude and the same hour angle ; but, unless n 
is very small, the two values will differ largely from each 
other, so that the observer has no dimculty in deciding which 
of them should be taken. 



whence, 



APPLICATIONS 



19; 



SECTION LXXIII 

PROBLEM 

Given the declination, the right ascension of a star, and the 
obliquity of the ecliptic, to find the latitude and the longitude of 
the star. 

Let M (Fig. 121) be the star, P the pole of the celestial 
equator, and Q the pole of the ecliptic. 

Then, in the triangle PMQ, 



given 


PQ = e = 


2,3° 27', . 








PM = 90° - d, 








ZMPQ = 90° + r; 






required 


QM= 90°- u, 






and 
where 


Z PQM = 90° - v ; 

r = the right ascension = TV?, 
u = the latitude of M = MT, 






and 


v = the longitude of M = VT. 


Q PH 




Illlllli 


% l '''- ■ 






llll! r' 


^i-^^ 7 ^^ 








aV' w 


■ :^-=^ / 


J M\ 


\f 


p&^ 


:-J^B 






"^J 5 


: ^ 


^^V R 






: -\ "'^I^W 




Fig. 121 






Fia. 120 







Here, two sides and the included angle are given. Draw 
MH J_ to PQ, and meeting it produced at H. Let PH = n. 



196 



SPHERICAL TRIGONOMETRY 



PH 




By Napier's Rules, 



sin r = tan n tan d, (1) 

sin u = cos (e -f- ri) cos Mff, (2) 
sin d = cos rc cos MH, (3) 

sin (e + n) = tan t? tan MH, (4) 

sin w = tan r tan ilfiiT. (5) 

From (1), 

FlG ' 122 tan n = cot d sin r. (A) 

From (3), cos MH = sin ^ sec n. 
Substitute in (2) the value of cos MH, 

sin u = cos (e + n) sin d sec n. (B) 

From (4), tan v = sin (e + n) cot Mff. (6) 

From (5), cot MH = tan r esc w. 
Substitute in (6) the value of cot MH, 

tan ?; = sin (e + w) tan r esc n. (C) 

Equations (A), (B), and (C) determine u and v. 
To avoid obtaining u from its sine we may proceed as 
follows : 

From equations (B) and (C) we have, by division, 
sin u _ sin d cos (e + n) sec w, 
tan v tan r sin (e -f- w) esc w 
.". sin u = tan v sin c? cot r cot (e + ?a) tan n. 

By taking Jl/fl" as middle part, successively, in the triangles 
MQH and MPH, we obtain 

cos MH = cos. u cos v, 

and cos il/HT = cos d cos r. 

. ' . cos u cos v = cos d cos r. 

.'. cos m = sec v cos e£ cos r. 
From these values of sin u and cos u we obtain, by division, 



APPLICATIONS 197 

sin u tan v cot (e + n) sin d cot r tan n 
cos m sec t; cos d cos /■ 

.'. tanw = sin v cot (e + n)tan d esc r tan n. 
From the relation 

sin r — tan ?i tan d, 
it follows that, dividing by sin /•. 

tan d esc r tan 91 = 1. 
Therefore, tan u = sin v cot (e -f w), 

a formula by which u can be easily found after v has been 
computed. 

EXERCISE XLIV 

1. Find the dihedral angle made by adjacent lateral faces 
of a regular ten-sided pyramid ; given the angle V == 18°, made 
at the vertex by two adjacent lateral edges. 

2. Through the foot of a rod which makes the angle A with 
a plane a straight line is draws in the plane. This line makes 
the angle B with the projection of the rod upon the plane. 
What angle does this line make with the rod ? 

3. Find the volume V of an oblique parallelopipedon ; given 
the three unequal edges a, b, c, and the three angles /, m, ??, 
which the edges make with one another. 

4. The continent of Asia has nearly the shape of an equi- 
lateral triangle, the vertices being the East Cape, Cape Romania, 
and the Promontory of Baba. Assuming each side of this 
triangle to be 4800 geographical miles, and the earth's radius 
to be 3440 geographical miles, find the area of the triangle : 
(i) regarded as a plane triangle ; (ii) regarded as a spherical 
triangle. 

5. A ship sails from a harbor in latitude I and keeps on the 
arc of a great circle. Her course (or angle between the direction 



198 SPHERICAL TRIGONOMETRY 

in which she sails and the meridian) at starting is a. Find 
where she will cross the equator, her course at the equator, and 
the distance she has sailed. 

6. Two places have the same latitude I, and the distance 
between the places, measured on an arc of a great circle, is d. 
How much greater is the arc of the parallel of latitude between 
the places than the arc of the great circle ? Compute the 
results for I = 45°, d = 90°. 

7. The distance d between two places and the latitudes I 
and I' of the places are known. Find the difference between 
their longitudes. 

8. Given the latitudes and longitudes of three places on 
the earth's surface, and also the radius of the earth ; show 
how to find the area of the spherical triangle formed by arcs 
of great circles passing through the three places. 

9. The distance between Paris and Berlin (the arc of a 
great circle) is equal to 472 geographical miles. The latitude 
of Paris is 48° 50' 13"; that of Berlin, 52° 30' 16". When 
it is noon at Paris what time is it at Berlin ? 

Note. Owing to the apparent motion of the sun, the local time over 
the earth's surface at any instant varies at the rate of one hour for 15° of 
longitude ; and the more easterly the place, the later the local time. 

10. Given the altitude of the pole 45°, and the azimuth 
of a star on the horizon 45°; find the polar distance of the 
star. 

11. Given the latitude I'of the observer, and the declination 
d of the sun ; find the local time (apparent solar time) of sun- 
rise and sunset, and also the azimuth of the sun at these times 
(refraction being neglected) . When and where does the sun rise 
on the longest day of the year (at which time d = -f- 23° 27') 
in Boston (I = 42° 21'), and what is the length of the day from 
sunrise to sunset ? Also, find when and where the sun rises 



APPLICATIONS 199 

in Boston on the shortest day of the year (when d = — 23° 27'), 
and the length of this day. . 

12. When is the solution of the problem in Example 11 
impossible, and for what places is the solution impossible ? 

13. Given the latitude of a place and the sun's declination ; 
find his altitude and azimuth at 6 o'clock a.m. (neglecting 
refraction). Compute the results for the longest day of the 
year at Munich (I = 48° 9'). 

14. How does the altitude of the sun at G a.m. on a given 
day change as we go from the equator to the pole ? At what 
time of the year is it a maximum at a given place ? (Given 
sin h = sin I sin d.) 

15. Given the latitude of a place north of the equator, and 
the declination of the sun ; find the time of da}' when the sun 
bears due east and due west. Compute the results for the 
longest day at St. Petersburg (I = 59° 56*). 

16. Apply the general result in Example 15 (cos t = cot I 
tan d) to the case when the days and nights are equal in length 
(that is, when d = 0°). Why can the sun in summer never be 
due east before 6 a.m., or due west after 6 p.m.? How does 
the time of bearing due east and due west change with the 
declination of the sun ? Apply the general result to the cases 
where I < d and I = d. What is it at the north pole ? 

17. Given the sun's declination and his altitude when he 
bears due east ; find the latitude of the observer. 

18. At a point in a horizontal plane MN a, staff OA is 
fixed so that its angle of inclination A OB with the plane is 
equal to the latitude of the place, 51° 30' N., and the direction 
OB is due north. What angle will OB make with the shadow of 
OA on the plane, at 1 p.m., when the sun is on the equinoctial? 

19. What is the direction of a wall in latitude 52° 30' K 
which casts no shadow at 6 a.m. on the longest day of the year? 



200 SPHERICAL TRIGONOMETRY 

20. Find the latitude of the place at which the sun rises 
exactly in the northeast on the longest day of the year. 

21. Find the latitude of the place at which the sun sets at 
10 o'clock on the longest day. 

22. To what does the general formula for the hour angle, 
in Sect. LXX, reduce when (i) h = 0°, (ii) I = 0° and d = 0°, 
(iii) I or d = 90° ? 

23. What does the general formula for the azimuth of a celes- 
tial body, in Sect. LXXI, become when t = 90° = 6 hours ? 

24. Show that the formulas of Sect. LXXII, if t = 90°, lead 
to the equation sin I = sin h esc d ; and that if d = 0°, they lead 
to the equation cos I = sin h sec t. 

25. Given the latitude of the place of observation 52°30'16", 
the declination of a star 38°, its hour angle 28° 17' 15"; find 
the altitude of the star. 

26. Given the latitude of the place of observation 51° 19' 20", 
the polar distance of a star 67° 59' 5", its hour angle 15° 8' 12"; 
find the altitude and the azimuth of the star. 

27. Given the declination of a star 7° 54', its altitude 
22° 45' 12", its azimuth 129° 45' 37"; find the hour angle of 
the star and the latitude of the observer. 

28. Given e = 23° 27' and the longitude y of the sun ; find 
the declination d and the right ascension r. 

! ' 29. Given e — 23° 27', the latitude of a star 51°, its longi- 
tude 315° ; find its declination and its right ascension. 

30. Given the latitude of the observer 44° 50', the azimuth 
of a star 138° 58', its hour angle 20°; find its declination. 

31. Given the latitude of the place of observation 51° 31 '48", 
the altitude of the sun west of the meridian 35° 14' 27", its 
declination + 21° 27'; find the local apparent time. 

32. Given the latitude of a place I, the polar distance p of 
a star, and its altitude h ; find its azimuth a. 



FORMULAS 

PLANE TRIGONOMETRY 

1. sin 2 A + cos 2 A = 1. 
sin y1 



2. tan .4 



cos .4 



sin A x esc A = 1. 

3. «! cos y! x sec A = 1. 
[tan J. x cot A = 1. 

4. 'sin (a; + ?/) = sin x cos 3/ + cos x sin y. 

5. cos (x + 2/) = cos & cos ?/ — sin cr sin ?/. 

tan x + tan y 

6. tan (# + ?/) = — — • 

x ' 1 — tan x tan y 

cot x cot y — 1 

7. cot(z + ?/) = — 7 r^T 

v 7 cot y -f cot # 

8. sin (x — y)— sin cc cos y — cos # sin y. 

9. cos (x — y) — cos # cos ?/ + sin x sin ?/. 

tan # — tan y 

10. tan (x-y)=— — — *-■ 

v J 1 + tan # tan y 

cot # cot y 4- 1 

11. cot(x-y) = ^-7- — 

v ' cot ?/ — cot X 

12. sin 2 a? = 2 sin a; cos x. 

13. cos 2 a; = cos 2 x — sin 2 sc. 

. ■ 2 tan jc 

14. tan2sc = ; — r— 

1 — tannic 

202 



15. cot 2x = 



FORMULAS „ 203 

cot 2 a: — 1 



2 cot x 
16. sin \ z = ± 



VI — cos z 



/l 4- cos « 
17. cosi^ = ±^/ s 



ll — cos z 
18. tanls = ±\ T - 

2 * 1 + cos 2 



-*& 



i9. •cot^g = ±M : ' • 

cos z 

20. sin A + sin 5 = 2 sin J (4 + 5) cos £ (.1 - B). 

21. sin .1 - sin B = 2 cos £ (.1 + i>') sin } {A - B). 

22. cos A + cos B = 2 cos £ (.1 + £) cos J(A — B). 

23. cos ,1 - cos B = - 2 sin £ (.1 + B) sin J (.1 - B). 

sin 4 -f- sin B tan £ (J + B) 
sin ^L — sin B tan i (.1 — B) 



« 



& sin 1> 
26. a 2 = & 2 + c 2 -2&ccos4. 

a — 6 tan J- (.1 - S) 



27 



a + b tan i (.1 + B) 



28. sini 



2 " * &< 



29. cos^=#Z5. 



204 FORMULAS 



30. 


-i-V^^r 1 


31. 


j(s-a)(s-b)(s-e) 


v 


32. 


r 

tan \A = 

s — a 


33. 


F =\ac sin B. 


34. 


a 2 sin B sin C 

x^ — . - • 



2 sin (B + C) 
35. F = Vs (s - a) (* - ft) (* - c). 

4i£ 

37. F=£r(a + £ + c) = ™- 

SPHERICAL TRIGONOMETRY 

38. cos c = cos <x cos 6. 
f sin a = sin c sin A. 



sin & = sin c sin 5. 



f cos A = tan & cot c. 

40. -4 

1 cos B = tan a cot c. 

f cos t! = cos a sin 5. 

41. -^ 

^ cos 5 = cos 5 sin ^. 

f sin b = tan a cot A. 
1 sin a = tan b cot I?. 

43. cos c = cot .4 cot 5. 

f sin a sin 5 = sin b sin A 

44. W sin a sin C = sin c sin 4. 

! sin b sin C = sin c sin i?. 



FORMULAS 205 



f cos a = cos ft cos c + sin & sin c cos 4. 

45. < cos & = cos a cos c + sin a sin c cos B. 

[ cos c = cos a cos ft 4- sin a sin 6 cos C. 

( cos 4 = — cos B cos C + sin 5 sin C cos a. 

46. ^ cos B = — cos A cos C + sin A sin C cos ft. 

I cos C = — cos 4 cos 2? + sin .4 sin i? cos c. 



47. 



sin \ A = V sin (s — ft) sin (s — c) esc ft esc c. 

cos ^ A = V sin s sin (s — a) esc ft esc c. 

tan J A = Vcsc s esc (s — a) sin (5 — ft) sin (s — c). 



f sin ^ a = V— cos S cos (.V — A) esc 5 esc C. 
48. -^ cos \a = Vcos (5 — 7?) cos (S — C) esc B esc C. 

[ tan \a = V— cos S cos (S — A) sec (5 — B) sec (if — C). 

f cos i- (.1 + B) cos £ c = cos I (a + b) sin J C. 

I sin I (.1 4- 22) cos ^ c = cos \ (a — ft) cos \ < '. 

1 cos i (A — 2J) sin 1 c = sin ^ (a + ft) sin J C. 

I sin 1 (.1 — 23) sin 1 c = sin \ (a — ft) cos £ C 

2 v ' cos i- (<x 4- ft) J 

tanK-l-B)^ 5in f^7^ cot^. 
2 v y sin i (a 4- ft) 

cos i(4 — B) x , 

sini(.4 — £), 
tan 1 (a — ft) = - — 2 ; ( tan i c. 

^ v y sm| (.4 4- B) 2 

M „ itR 2 E 
51 - F =-180- 

52. tan 2 i E = tan 1 5 tan \(s — a) tan 1 (s — ft) tan \ (s — c). 



50. ^ 



206 



FORMULAS 



The following diagram shows Prof. Blakslee's construction 
by which the direction ratios for plane right triangles give 
directly from a figure the analogies for a right trihedral or for 
a right spherical triangle : 



B, 




Fig. 123 

The construction consists of two parts. 

1. Lay off from the vertex V a unit's distance on each edge. 

2. Pass through the three extremities of these distances 
three planes perpendicular to one of the edges, as VA. ISTow 
these three parallel planes will cut out three similar right 
triangles. The first being constructed in either of the two 
usual ways, the construction of the others is evident. 

Since the plane angles A±, A 2 , A z all equal the dihedral A, 
and the nine right triangles in the three faces give the values 
in the figure, we have : 

(1) sin A = sin a : sin h ; similarly, sin B = sin b : sin h. 

(2) cos A = tan b : tan h ; similarly, cos B = tan a : tan h. 

(3) tan A = tan a : sin b ; similarly, tan B = tan b : sin a. 

(4) cos h = cos a cos b ; by (3), cos h = cot A cot B. 

(5) sin A =? cos B;, cos b ; sin B = cos A : cos a. 



FORMULAS 207 

Note. If a sphere of unit radius is described about Fas a centre, 
the three faces will cut out a right spherical triangle, having the sides a, 
6, and A, and angles A, B, and H. The above formulas are thus seen to 
be the analogies of : 

(1) sin A — a : h ; sin B = b : h. 

(2) cos A = b : h ; oos B = a : h. 

(3) tan A = a : b ; tan B = b : a. 

(4) h' 2 - a 2 + b 2 ; 1 = sin 2 + cos 2 ; 1 = cot A cot B. 

(5) sin A = cos B ; sin I? = cos .A. 

Napier's Rules give only the following, which follow from the analogies 
as numbered : 

By j sin a = sin ^4 sin h = tan 6 cot J? i 
(1) I sin b — sin i? sin h = tanacot^L J 
r cos J. = sin B cos a = tan b cot A ^ 
^ I cos J5 = sin A cos 6 = tan a cot h ) 
(4) { cos /i = cos a cos b — cot J. cot B } (4) 



GAUSS'S EQUATIONS 

cos ^ (J + B) cos ^ c = cos £ (a + 6) sin £ C. 
sin i (.4 -f I?) cos i- c = cos ^ (a — b) cos J- C. 
cos J (J — I?) sin ^ c = sin i (a 4- 6) sin ^ C. 
sin i (.4 — 23) sin J c = sin J (a — b) cos J C. 



SURVEYING 

CHAPTER I 
FIELD INSTRUMENTS 

SECTION I 
DEFINITIONS 

Definition. Surveying is the art of determining and repre- 
senting distances, areas, and the relative position of points on 
the surface of the earth. 

Classification. Of surveying there are various kinds, depend- 
ing upon the extent, the purpose, or the method of the survey. 
The following are the principal divisions : 

1. Plane Surveying, in which the part of the earth's surface 
surveyed is regarded as a plane ; Geodetic Surveying, in which 
the true figure of the earth is regarded. 

2. Land Surveying, in which boundary lines, contents, and 
outline maps are the chief things aimed at; Topographic Sur- 
veying, in which differences in elevation and contour maps are 
chiefly sought ; Hydrographic Surveying, in which the purpose 
is to determine the configuration and topography of the bed or 
basin of a body of water ; Mine Surveying, in which the posi- 
tion and extent of underground excavations are determined and 
graphically represented. 

3. Rectangular Surveying, in which a system of perpen- 
dicular lines is used as reference lines ; Triangular Surveying, 
which proceeds by means of a system of triangles referred to 
a well established base line. 

209 



210 SURVEYING 

Operations Comprised. Surveying commonly comprises the 
following three distinct operations : 

1. The Field Measurements, or the determining certain lines 
and angles by direct measurement. 

2. The Computation of the required parts from the meas- 
ured lines and angles. 

3. The Plotting, or representing on paper the measured and 
the computed parts in relative extent and position. 

Historic Note. Surveying is undoubtedly one of the oldest of the arts 
of civilized man. The Bible contains several admonitions not to 
remove " the ancient landmark," as in Proverbs xxii. 28. To the Baby- 
lonians is credited the division of the circle into 360 degrees. The Egyp- 
tians were known to survey frequently the valley of the Nile, a necessity 
owing to the periodic overflow of that river. Thence came Geometry. The 
Egyptians also possessed rules for finding the area of land of various shapes. 
Moreover, on Egyptian soil the Greek mathematician Eratosthenes made 
the first attempt at determining the circumference of the earth by meas- 
uring an arc of the circumference. This was in 276 b.c. Among the 
Romans Surveying was considered one of the liberal arts, and received 
impetus in the time of Julius Caesar from his sweeping order that the 
entire empire should be surveyed for the purpose of equitable adjustment 
of taxes, and also from the introduction of the more practical parts of 
Greek Geometry. The works of Roman surveyors served as models for 
centuries, and much that we have to-day is only improvements on what has 
been handed down from them. For a brief account of surveying in the 
United States, see Cajori's "The Teaching and History of Mathematics 
in the United States," pp. 92, 286. 



FIELD INSTRUMENTS 211 

SECTION II 
THE CHAIN 

Surveyor's Chain. The Surveyor's Chain, or Gunter's Chain 
as it is often called, is made of iron or steel wire and is 4 
rods or 66 feet long, composed of 100 links connected by small 
rings, and provided with a tally mark at the end of every 10 
links. A link as a unit of measure includes a link of the chain 
and half the rings that connect it with adjoining links. Each 
link is 7.92 inches long. Since a chain is 4 rods long, a square 
chain contains 16 square rods, and since an acre contains 160 
square rods, a square chain is one-tenth of an acre. A square 
chain contains also 10,000 square links and, therefore, an acre 
contains 100,000 square links. Hence, if a given area is 
expressed in square chains, it is reduced to acres by pointing 
off the last figure, and, if expressed in square links, it is reduced 
to acres by pointing off the last five figures. The tally marks 
are appropriately notched to facilitate counting links from 
either end, the one at the middle of the chain being rounded 
so as to be distinguished readily from the others. Handles 
form part of the end links, to which they are so attached as 
to prevent twisting and to allow lengthening or shortening 
of the chain. The Surveyor's Chain is used in measuring 
land. 

Engineer's Chain. The Engineer's Chain differs from the 
ordinary Surveyor's Chain chiefly in that it is 100 feet in 
length, the length of each link being 1 foot. It is used in 
surveying railroads and canals, and often in other surveys 
where extensive lines are to be run. 

Both the Surveyor's Chain and the Engineer's Chain are 
generally provided with attachments, so that from the full 
chains half-chains can be made up, to be used in case of rough 
or hilly country. 



212 SURVEYING 

Accompanying Pieces. Usually eleven, sometimes ten, Mark- 
ing pins go with the chain. These are of iron or steel, about 
14 inches long, pointed at one end and formed into a ring at 
the other end. In case eleven pins are used, the first one is 
placed at the beginning of the line to be measured, and there- 
after one at the end of each chain. The last pin in the ground 
is, therefore, not to be counted. In case ten pins are used, 
the first one is placed at the end of the first chain, and so on, 
the last pin in the ground being counted. Strips of red cloth 
should be fastened to the ring ends of the pins so as to make 
them easily visible. Ranging poles, which are of various 
lengths, are necessary for alignment. These are commonly 
made of wood, and are steel shod, graduated to feet, and 
painted in alternate red and white stripes. 

How to chain. Ranging poles should be placed, one at each 
end of the line to be measured, and at such intermediate 
points as the necessities of the case require. A head chain- 
man or leader, and a rear chainman or follower are required. 
The follower takes one end of the chain, and one pin, which 
he thrusts into the ground at the beginning of the line. The 
leader takes the other end of the chain and the remaining ten 
pins, and moves forward until the word " Halt " from the fol- 
lower warns him that he has advanced nearly the length of 
the chain. At this signal he stops, and the follower, mean- 
while having placed his end of the chain against the pin at 
the beginning of the line, directs the leader by the words 
"Right" and "Left" until he is exactly in the line. This 
being accomplished, and the chain tightly stretched in a hori- 
zontal position, the follower says, " Down." The leader then 
puts in a pin at the end of the chain and answers, " Down " ; 
after which the follower withdraws the pin at his end of the 
chain, and the chainmen move forward, repeating the process 
just described until the end of the line is reached. 

If the marking pins in the hands of the leader are all placed 



FIELD INSTRUMENTS 213 

before the end of the line is reached, after putting the last pin 
in the ground he waits until the follower comes up to him, 
gives him the ten pins in his hands and records the fact that 
ten chains have been measured. The measuring then proceeds 
as before. If the distance from the last pin to the end of the 
line is less than a chain, the leader places his end of the chain 
at the end of the line, and the follower stretches tightly such 
part of the chain as is necessary to reach the last pin, and 
the number of links is counted. If the ground slopes, one end 
of the chain must be raised until the horizontal position is 
attained. By means of a plumb line or a slender staff or, less 
accurately, in case of the leader by dropping a pin (heavy end 
downwards), the point vertically under the raised end of the 
chain may be determined. If the slope is considerable, half 
a chain or less may be used ; in which case care must be taken 
that the correct number of full chains and links is found. For 
instance, if a tally shows 15 half chains and 35 links, the 
appropriate measure is 7 chains and So links, or, as it is 
usually expressed, 7.85 chains. 

Special Constructions by Means of the Chain. 1. At a given 
point in a given line to construct a perpendicular to that 
line. 

Let LE (Fig. 1) be the given line, and P the given point. 
On LE measure off PB = PA = 20 links. Then place one end of 
the chain at B and the other end at A. 
Stretch the chain from the middle point, / \ 

and mark that point, as C. PC is the / 

perpendicular required. (Why ?) / 

Or, make PB = 30 links. Place one - — £ 

end of the chain at P, and the end of 

the 90th link at B. Then, taking the 

chain at the end of the 40th link from P and stretching 

both portions tightly, mark that point, as C. Then PC is the 

perpendicular required. (Why ?) 



E 



P 
Fig. l 



214 



SURVEYING 



2. Through a given point without a given line to construct 
a perpendicular to that line. 

Let LE (Fig. 1) be the given line, and C the given point. 
Take any point as B in the line and stretch the chain between 
C and B ; then swing the chain about C until the point at B is 
again in the line, as at A. Measure the distance between A 
and B. Then P, the mid-point of AB, is a second point in the 
required perpendicular. (Why ?) 

Or, let the middle of the part of the chain between C and B 
be held in place, and swing the end at C until it meets the 
line as at P. PC is the required perpendicular. (Why ?) 

3. At a given point in a given line to construct an angle 
equal to a given angle. 





Let P (Fig. 2) be the given point in the given line LE, 
and angle A the given angle. Make PD = AB. At D and B, 
respectively, construct perpendiculars BF and BC. Make 
DO = B.C. Then angle OPD is the angle required. (Why ?) 

4. To construct any given angle, as 25° 40'. 

Find from the tables the tangent of 25° 40', which is 0.4806. 
Lay off PD (Fig. 2) = 100 links. Construct the perpendicular 
DF and lay off DO = 48.06 links. Then angle OPD is the 
angle required. (Why ?) 

5. Through a given point to construct a line parallel to a 
given line. 

Let P (Fig. 3) represent the given point, and LE the 
given line. Through P lay out any convenient line as BA 



FIELD INSTRUMENTS 



215 



intersecting LE. Construct angle BPD = angle PAE. Then 
the line CD is the required line. ' (Why ?) 



/B 



1L 



D 



A/ 




II 



D 



Fig. 4 



Fig. 3 

Obstacles to chaining. In general practice various obstacles 
are encountered in chaining. The circumstances in each case 
must decide the best method to be used. Only a few sugges- 
tive cases can be considered in this work. 

1. To measure a line when a building, or other object, 
stands in the way. 

In Fig. 4 construct the 

perpendicular AB, the per- -^ 

pendicular BC, the perpen- 
dicular CD = AB, then the 
perpendicular DE, which 
will be in line LA prolonged. 

Then, LA + BC + DE = LE. (Why ?) As a check, another 
series of perpendiculars may be constructed. 

2. To measure across a body of water. 

At A (Fig. 5) lay 

out AP, making angle 
PAB = 60°. This can 
be done by laying out 
the equilateral tri- 
angle A BD. At P 
range out PC, mak- 
ing angle APC = 60°. 
Then measure A P. 




r# 



Fig. 5 



216 



SURVEYING 



The line AC is equal to AP. (Why ?) If C is some fixed 
point in LE, across the stream, accessible or inaccessible, we 
may proceed as follows : After laying out AP, as already 
described, with 90 links of the chain stretched in the form 
of an equilateral triangle, and with one side of this triangle 
in AP, move the triangle until the point C is in line with 
the forward side of the triangle. Then proceed as before. 

3. To measure a line the end of which is invisible from 
the beginning, and the intermediate points are unknown. 



D! 



B ~~~~-~- 



—R 



Fig. 6 



Let LE (Fig. 6) represent the line. Lay out the line LR so 
that R shall be beyond E and visible from L. Construct 
from E the perpendicular EA to LR. Measure LA and AE. 
LE can then be computed. (How ?) If intermediate points 
on LE are to be sought, take any point in LA, as B ; construct 
BC perpendicular to LA ; then measure off BD of such length 
that BD : AE = LB: LA. The line LR is called a random 

line. 
4 



To measure the dis- 
tance between two inaccessi- 
ble points. 

Let L and E (Fig. 7) be 
two inaccessible points. 
Select some point as P from 
which both L and E are 
visible. Measure PL and PE 




PX 



E' 



Fig. 7 



by the method in 2. Range 



FIELD INSTRUMENTS 217 

out PL' in line with LP and equal to LP ; similarly, PE' = EP. 
Then measure L'E'. which is equal to LE. (Why?) 

EXERCISE I 

1. Range out a line which, by estimation, is more than 
10 chains long. Then measure it with the chain out and 
back. 

2. Prolong a line beyond a building, or other obstacle which 
prevents continuous alignment. 

3. Find the distance from a point to a line when the dis- 
tance is more than a chain. 

4. Lay out a square field each side of which shall be 5.76 
chains long. 

5. Find the length of a line by means of a random line. 
Then, as a check, find its length by direct measurement. 



SECTIOX III 
THE TAPE 

Kinds of Tape. The tape measure used by the surveyor 
or engineer consists of a thin ribbon of steel, or of linen 
with interwoven wires of brass, wound upon a reel, often in 
a leather or metal case. Tapes vary in length from 25 feet 
to 500 feet or more. They are variously graduated to links, 
to feet and inches, to feet and tenths of a foot, to metric units, 
or to a combination of these. A common combination is feet 
and tenths of a foot on one side, and links on the reverse side. 

Uses. The kind of tape determines to a great extent the 
use to which it is to be put. If 33 feet or 66 feet long and 
graduated to links, the evident purpose is for land survey- 
ing. If 50 feet or 100 feet long and graduated to feet and 



218 



SURVEYING 



tenths of a foot, the tape is especially designed for city work. 
Other kinds are employed in bridge, road, or mining work, in 
very accurate measurements of base lines, or as standards of 
comparison for other instruments of measurement. 



SECTION IV 
THE COMPASS 

Parts and their Uses. The essentials of the compass, one 
style of which is shown in Fig. 8, are : the compass circle, 
graduated to half degrees and figured from 0° to 90° each way 




Fig. 



The Surveyor's Compass 



Note. The letters E and W on the face of the compass are reversed 
from their true positions. The reason for this is that if the sights are 
turned towards the west, the north end of the needle is turned towards 
the letter W, and if the north end of the needle is turned towards E, the 
sights are turned towards the east. 

If the north end of the needle points exactly towards E or W, the 
sights range east or west. 



FIELD INSTRUMENTS 219 

from the north and south points, for indicating the directions 
of lines ; the magnetic needle, pivoted on a pin at the centre 
of the compass circle, for showing the direction of the mag- 
netic meridian ; and the sight standards, attached to the ends 
of the main plate, for alignment. To the main plate are 
attached two spirit levels at right angles to each other for 
leveling the instrument; underneath is a needle-lifting screw 
which, by actuating a concealed spring, lifts the needle from 
the pivot and presses it against the glass covering of the 
compass circle when the instrument is not in use ; a tangent 
screw, and almost directly under it a clamp screw, which 
operates the vernier; and a small dial plate for keeping tally 
in chaining. The north end of the needle usually has some 
ornamentation to distinguish it from the south end, and a 
coil of tine wire is wound on the south end to prevent the 
needle from dipping. The sight standards have fine slits 
nearly their whole length, with circular openings at intervals 
to facilitate sighting upon an object ; on the edges of the north 
standard are tangent scales for reading vertical angles, and 
on the outside of the south standard are two eyepieces at the 
same distance from the main plate as the zeros of the tangent 
scales, respectively. The telescopic sight is an attachment to 
the south standard, now often used. The instrument entire 
turns horizontally upon the upper end of a ball spindle, the 
lower end of which rests in a spherical socket in the top of 
a Jacob's staff, or a tripod, which supports the instrument. 
The socket of the compass which fits to the ball spindle is 
provided with a clamp screw and a spring catch. From the 
centre of the plate at the top of the tripod a plummet is 
suspended by which the centre of the compass can be placed 
directly over a definite point on the ground. 

Kinds of Compasses. The compass described is the vernier 
compass, or surveyor's compass, and is the one in general use. 
If there is no vernier attachment, the compass is called & plain 



220 



SURVEYING 



compass and is used in running new lines and the preparation 
of maps. A railroad compass has all the features of the vernier 
compass, and has also a vernier plate and graduated limb for 
measuring horizontal angles. 

Hints on the Use and Care of Instruments. The instruments 
described in this work are adjusted by the maker. If they 
should require readjustment, full directions will be found 
in the manual furnished with the instruments. Before begin- 
ning to use any instrument, make a thorough study of its 
various parts and their uses. In moving or adjusting any 
part always know what you are doing and why you are doing 
it. When an instrument is not in use keep it in a place that 
is free from moisture and dust. 

Bearing of a Line. The magnetic meridian of a place is the 
direction which a bar magnet assumes when freely supported 

in a horizontal position. 
The magnetic bearing of a 
line is the angle it makes, 
with the magnetic merid- 
ian. To take the bearing 
of a line proceed as fol- 
lows : Place the compass 
so that the Jacob's staff, 
or plummet of the tripod, 
is directly over one end of 
the line, and level by press- 
ing with the hands on the 
main plate until the bub- 
bles are brought to the 
centres of the spirit levels. 
Turn the south end of the instrument toward you, and sight 
at the ranging pole at the other end of the line. Read the 
bearing from the north end of the needle. First, write 1ST. or 
S. according as the north end of the needle is nearer K or S. 




Fig. 9 



FIELD INSTRUMENTS 221 

of the compass circle ; secondly, write the number of degrees 
between the north end of the needle and the nearest zero mark ; 
thirdly, write E. or W. according as the north end of the needle 
is nearer E. or W. of the compass circle. Thus, in Fig. 9 (a), the 
bearing is N. 45° W. ; (b), K 60° E. ; (c), S. 60° W. ; (d), S. 45° E. 

If the needle coincides with the N.S. or E.W. line, the 
bearing is 1ST., S., E., or W. according as the north end of the 
needle is over IN"., S., E., or W. As the compass circle is 
divided into half degrees, the bearing may be determined 
pretty accurately to quarter degrees. 

It will be noticed that the letters E and W on the face of 
the compass are reversed from their true positions. These are 
so placed in order that when the sights are turned towards 
the west the north end of the needle will point towards the 
letter W, or if the sights are turned towards the east, the 
north end of the needle will point towards the letter E. It 
turns out that if the south sight standard is always turned 
towards the observer, the reading at the north end of the 
needle will indicate the true bearing of the line. Should the 
north sight standard be turned towards the observer, the read- 
ing at the south end of the needle would then be taken. 

Checking Bearings. When the bearing of a line has been 
taken, the instrument should be removed to the other end of 
the line and the reverse bearing taken. The number of 
degrees should be the same, but the letters should be reversed. 
For instance, if the direct bearing is N. 35f° W., the reverse 
bearing should be S. 35f ° E. In case the reverse bearing is 
not what it ought to be, there is some mistake, or some local 
disturbance, or both. To detect errors a second trial at the 
direct bearing should be taken. To detect local disturbances 
take the direct and reverse bearings of other lines ranged out 
from the beginning of the line whose bearing is sought. If 
they all show the same difference between their two respective 
bearings, the evidence of some local disturbance, as iron, 



222 SURVEYING 

iron ore, etc., is pretty conclusive. In this case the true bear- 
ing of the line can be obtained by making the necessary cor- 
rection. In all cases, precautions should be taken to have the 
chain, pins, and other instruments that would affect the direc- 
tion of the needle sufficiently removed from the compass. 

Obstacles. When a fence or other obstruction interferes 
with placing the instrument over the line the instrument may 
be placed at one side, the ranging pole being correspondingly 
placed at the other end. If one end of the line cannot be seen 
from the other end, run a random line. Then (Fig. 6, p. 216) 
tan EL A = AE -r- LA, whence the angle ELA can be found. 
This angle combined with the bearing of the random line will 
give the bearing required. Or some point can be selected 
from which the ends of the line are visible. The distances to 
the ends may be measured, and the angle between the two 
auxiliary lines may also be measured. Of the triangle thus 
formed, the angle at the beginning of the given line may be 
computed, and, when properly combined with the bearing of the 
first auxiliary line, will give the required bearing. If a single 
triangle is not sufficient, a series of triangles may be employed 
until the end of the line is reached. 

Measurement of Horizontal Angles. To measure a horizontal 
angle by means of the needle, place the compass over the 
vertex of the angle, take the bearing of each line separately, 
and combine these bearings according to the following rules, 
as suggested by Fig. 10 : 

1. If the first letters of the bearings are alike, and also the 
last letters, find the difference of the bearings. 

2. If the first letters are alike, and the last letters unlike, 
add the bearings. 

3. If the first letters are unlike, and the last also unlike, 
subtract the difference of the bearings from 180°. 

4. If the first letters are unlike, and the last alike, subtract 
the sum of the bearings from 180°. 



FIELD INSTRUMENTS 



223 




Fig. 10 



Measurement of Vertical Angles. A vertical angle is an 
angle the sides of which are in a vertical plane. If one side 
of a vertical angle is horizontal and the other ascends, the 
angle is called an angle of elevation ; if one side is horizontal 
and the other descends, the angle is called an angle of depres- 
sion. To measure an angle of elevation by means of the com- 
pass, sight through the lower eyepiece to a point that is as 
far above the point whose elevation is sought as the instru- 
ment is above the point from which the elevation is to be 
taken. Read off the degrees of the right-hand tangent scale, 
marked by a card placed squarely across the face of the south 
standard, the top of the card being in the line of sight. To 
measure an angle of depression, proceed in the same manner, 
using the upper eyepiece and the left-hand tangent scale. If 
the compass is provided with a telescopic sight that has a 
vertical circle attachment, these should be used instead of 
the eyepieces and tangent scales. 

Verniers. A vernier is a contrivance for measuring portions 
smaller than those into which a line is divided. We shall 
describe two kinds. 

Let AB (Fig. 11) be a portion of a line graduated to tenths 
and hundredths of a foot. VR is the vernier. 

In (a), nine parts of the line are divided into ten equal parts 
on the vernier. Hence, a division on the vernier is less than 
a division on the line by the difference between T i 7 of a foot 
and j\ of t £q of a foot, or ^^ of a foot. Now, if the vernier 



224 



SURVEYING 



is moved so that 1 of the vernier coincides with 1 of the scale, it 
has moved over a space equal to yqVtt °^ a ^ 00 ^ ^ the vernier 
is moved so that 2 of the vernier coincides with 2 of the scale, 
it has moved over a space equal to y^^ of a foot ; and so on. 
In (b), 6 of the vernier coincides with 9 of the scale, which 
indicates that the zero of the vernier has moved past 3 of the 
scale a space 'equal to T q\ ^ of a foot. The reading, then, is 





&• 

f oo <o -4 e* c 


jvf 




Mil 




<«)/ .mum 


Ml' 


1 



I 



II I I I 1 1. 1.1 



El 



III: 



fo) I 1 o>, i 1 I I I 



a co n (b >o <t 



n i i i i i i i i 



xn 



&M 



n 



•^ ^ g? >H 



>Q -^t SO S* 



(d)(| I I 



-* <» 00 



fl 1, 1. 1. I. I I I I I 



3 



R 



SO >Q -4 <5Q g-t >^i Ol 00 N 



mum 



Fig. 11 



0.536 foot. This form of the vernier is known as the direct 
form, since the figuring on the vernier proceeds in the same 
direction as that on the scale. 

In (c), eleven parts of the line are divided into ten equal parts 
on the vernier. Hence, a division on the vernier is greater 
than a division on the line by the difference between t l of 
T y_ of a foot and T i ¥ of a foot, or T J^ of a foot. Now, if the 
vernier is moved so that 1 of the vernier coincides with 10 



FIELD INSTRUMENTS 



225 



of the scale, i.e., the end of the 6th tenth, the vernier has 
moved over a space equal to yoV o °f a ^°ot. If the vernier is 
so moved that 2 of the vernier coincides with 9 of the scale, 
the vernier has moved over a space equal to T ^%^ of a foot ; 
and so on. 

In (d), 6 of the vernier coincides with 7 of the scale, which in- 
dicates that the zero of the vernier has moved past 3 of the scale 
a space equal to t15 %q of a foot. The reading here is 0.636 foot. 

This form of the vernier is known as the retrograde form, 
since the figuring on the vernier proceeds in the opposite direc- 
tion from that on the scale. In either form the following 
rule for using and reading the vernier may be adopted : 

Move the vernier until its zero line, or index, is at the point 
to which the required measurement is to be taken; read the 
main scale to the nearest division below the index, and that 
number of the division line of the vernier which stands opposite 
a, line of the main scale. 




Fig. 12 



Compass Vernier and its Uses. Let LV (Fig. 12) repre- 
sent the limb of the compass graduated to half degrees, and 
W the vernier divided into thirty equal spaces, equal to 
twenty -nine spaces of the limb. Then, one space of the vernier 
is less thanone space of the limb by l'(=30' — ^ of 29x30'), 
and the reading may be obtained to single minutes. 



226 



SURVEYING 



In Fig. 12 the index, or zero, of the vernier stands between 
32° and 32° 30', and the line of the vernier marked 9 coincides 
with a line of the limb. Hence, the reading is 32° 9'. 

When the index moves from the zero line of the limb in a 
direction opposite to that in which run the numbers of the 
limb, the number of minutes obtained as above must be sub- 
tracted from 30' to obtain the minutes required. (Why ?) If, 
however, the vernier is made double, that is, if it has thirty 
spaces on each side of the zero line, it is always read directly. 
The usual form of the double vernier, shown in Fig. 13, has 




20 


25 30 25 20 


,,. 


5 zero 

V 


b 10 


1 1 




1 















1 




♦ 




Fig. 13 



only fifteen spaces on each side of the zero line. When the 
vernier is turned to the right less than 15' past a division line 
of the limb, read the lower figures on the left of the zero line 
at any coincidence ; if moved more than 15' past a division 
line of the limb, read the upper figures on the right of the 
zero line at any coincidence ; and vice versa. In this form of 
the double vernier it will be observed that the spaces on the 
vernier are larger than those on the limb, since the 30 equal 
spaces of the former are equal to 31 half -degree spaces of the 
latter. 



FIELD INSTRUMENTS 227 

The most important use of the vernier compass is in setting 
off the variation of the needle explained just below. If the 
variation of the needle at any place is known, by means of the 
vernier screw the compass circle may be turned through an arc 
equal to the variation. If the observer stands at the south 
end of the instrument, the vernier is turned to the right or left 
according as the variation is west or east. The compass now 
gives the bearings of the lines with the true meridian. 

In order to retrace the lines of an old survey, turn the 
sights in the direction of a known line and move the vernier 
until the needle indicates the old bearing. If no line is defi- 
nitely known, the change of variation from the time of the 
old survey will give the arc to be set off. 

Magnetic Declination. The magnetic declination, or varia- 
tion of the needle, at any place is the angle which the magnetic 
meridian makes with the true meridian, or north and south 
line. The variation is east or west, according as the north end 
of the needle lies east or west of the true meridian. Western 
variation is indicated by the sign +, and eastern by the sign — . 
The kinds of magnetic declination are put under three heads : 

1. Irregular variations, which are sudden deflections of 
the needle due to magnetic storms or other causes not well 
understood. 

2. Solar-diurnal variations, which in northern latitudes reach 
their farthest point east about 8 o'clock a.m., and their farthest 
point west about 2 o'clock p.m., varying from 5' in the winter 
in some localities to 20' in the summer in other localities. 

3. Secular variation, which is a change in the same direction 
for a period of years, then in the opposite direction for about 
the same time. 

It is not accurately known how long it takes a complete 
secular variation to run its course, but from data already 
obtained it seems probable that the period of time covered is 
not less than two and a half or three centuries. 



228 



SURVEYING 



th H 

35 


8 

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Halifax, N.S. . . 
Eastport, Me. . . 
Bangor, Me. . . . 
Provincetown, Mass. 
Portland, Me. . . 
Portsmouth, N.H. . 
Boston, Mass. . . 
Cambridge, Mass. . 
Quebec, Canada . 
Providence, R.I. . 
Hartford, Conn. 
New Haven, Conn. 
Burlington, Vt. 
Williamstown, Mass. 
Montreal, Canada . 
Albany, N.Y. . . 
New York, N.Y. . 
New Brunswick, N.J. 
Cape Henlopen, Del. 
Philadelphia, Pa. . 
Cape Henry, Va. . 
Ithaca, N.Y. 
Baltimore, Md. . . 
Williamsburg, Va. 
Harrisburg, Pa. 
Washington, D.C. 
Newbern, N.C. . . 
Buffalo, N.Y. . . 
Toronto, Canada . 



FIELD INSTRUMENTS 



229 



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Charleston, S.C. . 
Pittsburg, Pa. . . 

Erie, Pa 

Savannah, Ga. . . 
Cleveland, 0. . . 
key West, Fla. 

Sault Ste. Marie, Mich 
Cincinnati, 0. . . 
Grand Haven, Mich. 
Nashville, Tenn. . 
Michigan City, Ind. 
Pensacola, Pla. . . 
Chicago, 111. . . . 
Milwaukee, Wis. . 
Mobile, Ala. . . 
New Orleans, La. . 
St. Louis, Mo. . . 
Duluth, Minn. . . 
Galveston, Tex. 
Omaha, Neb. 
Austin, Tex. . . 
San Antonio, Tex. 
Denver, Col. 
Salt Lake City, Utah 
San Diego, Cal. 
Seattle, Wash. . . 
San Francisco, Cal. 
C. Mendocino, Cal. 



230 SURVEYING 

The agonic line, or line of no variation, is a line joining 
those places at which the magnetic meridian coincides with 
the true meridian. At the beginning of the present century 
this line crossed the United States in an irregular way from 
Michigan to South Carolina. It is gradually moving west- 
ward, so that the variation is increasing at places east of this 
line, and decreasing at places west of the line. East of this 
line the variation is westerly, and west of this line the varia- 
tion is easterly. Lines that join places of equal magnetic 
declination are called isogonic lines. 

Table of Magnetic Declination. On pp. 228, 229 will be 
found a table showing the variation in magnetic declination at 
different places in the United States and contiguous territory 
during the nineteenth century ; also the annual change for the 
epoch of 1900. 

EXERCISE II 

1. Lay out a field of five sides and take the bearings and 
measures of the sides in order, beginning at the most westerly 
point and going about the field clockwise. 

2. From the bearings obtained in Example 1 find the value 
of each of the interior angles. What is their sum ? 

3. Lay out the field the bearings and distances of whose 
sides are given in Example 1 of Exercise VI, p. 272. 

4. Kange out a line whose bearing is N. 38° 30' W., and at 
some point in this line range out another line making a right 
angle with it. What is the bearing of the second line ? 

5. Set up the compass at a spot near which there is known 
to be some local disturbance, as iron in a building, or an 
iron fence, and find the variation of the needle due to such 
disturbance. 



FIELD INSTRUMENTS 231 

SECTION V 

THE TRANSIT 

Surveyor's Transit. The transit is the most important 
instrument used in surveying. There are many modifications 
of it, each adapted to its own particular use. All, however, 
have about the same essential features. The one described 
here, and shown in Fig. 14, is the surveyor's transit, the one 
of most general use. The essential parts are the telescope 
with its axis and two standards, the circular plates with their 
attachments, the sockets upon which the plates revolve, the 
leveling head, and the tripod. Within the telescope are two 
fine cross wires, at right angles to each other, whose intersection 
determines the optical axis, or line of collimation, of the tele- 
scope. Under the telescope, and attached to it, is a spirit 
level by which horizontal lines may be run, or the difference 
of level between two stations be found. The axis of the tele- 
scope carries a vertical circle which measures vertical angles 
to single minutes by means of a vernier. The vernier plate, 
which carries the telescope and also the compass circle, has 
two verniers diametrically opposite to each other, and it moves 
entirely around the graduated limb of the main plate. The 
sockets are compound ; the interior spindle attached to the 
vernier plate turning in the exterior socket, when an angle is 
taken on the limb, but when the plates are clamped together 
the exterior socket itself, and with it the whole instrument, 
revolves in the socket of the leveling head. The transit is 
leveled by four leveling screws which pass through a plate 
firmly fastened to the ball spindle and rest in small sockets, 
these resting in turn on the upper side of the tripod plate. 
On the underside of this lower or tripod plate is an arrange- 
ment called a shifting centre, which enables the surveyor to 
change the position of the vertical axis horizontally without 



232 SURVEYING 

moving the tripod ; besides this there is, if specially ordered, 
a device called a quick-leveling attachment to bring the transit 
quickly to an approximately level position by the pressure of 
the hands after which the leveling screws are used. 

Uses. The transit may be used for all the purposes for 
which the compass may be used, but with much greater 
precision. The principal use, however, is in measuring hori- 
zontal angles by means of the graduated limb and verniers. 
It may be used, furthermore, in obtaining differences of level ; 
also, provided there is the attachment to the telescope known 
as the stadia, in measuring distances, especially over broken 
ground. A still further use, when the transit is supplied 
with what is known as a gradienter attachment, is in fixing 
grades as well as measuring distances. 

Getting the Transit Ready. The instrument should be set 
up so as to be firm, the tripod legs being pressed into the 
ground until the plates are as nearly level as can conveniently 
be done by this means. For the subsequent leveling turn the 
instrument until the spirit levels on the vernier plate are 
parallel to the vertical planes passing through opposite pairs 
of the leveling screws. Take hold of opposite screw heads 
with the thumb and forefinger of each hand, and turn both 
thumbs in or out as is necessary to bring the bubble to its 
proper place, the left thumb always moving in the direction 
that the bubble is to move. For precise work, in addition to 
leveling by the leveling screws, it is advisable to level the 
plates by the telescope level, as this is much more sensitive 
than the levels on the plate. In this operation the position 
of the level on the telescope must be observed over both sets 
of leveling screws, one half the correction being made by the 
axis tangent screw, the other half by the leveling screws. 
Before an observation is made with the telescope, the eye- 
piece should be focused by its pinion until the cross wires 
appear distinct; the object glass is then focused by its pinion 



FIELD INSTRUMENTS 233 



Fig. 14. The Surveyor's Transit 



234 SURVEYING 

until the object to be observed appears well denned. This 
latter process must be repeated when the distance to the 
object is changed. 

Measurement of Horizontal Angles. Place the instrument 
directly over the vertex of the angle, and level. Set the 
limb at zero by the tangent screw of the plates, and turn the 
telescope in the direction of one of the sides of the angle, 
directing it to the object by the tangent screw of the leveling 
head. Then unclamp the main plate and turn the telescope 
until it is in the direction of the other side of the angle, 
and read the angle by the verniers. The object of the two 
verniers on the vernier plate is to correct any mistakes that 
might arise from the want either of exact coincidence in the 
centres of the verniers and the limb or of exact graduations 
on the limb. The correct reading may be obtained by adding 
to the reading of one vernier the supplement of the reading of 
the other, and taking half their sum. 

Measurement of Vertical Angles. Direct the telescope to the 
object ; clamp, and read the angle indicated on the vertical 
circle by the vernier. The angle read will be an angle of 
elevation or depression as the case may be, the horizontal line 
being the line of collimation of the telescope when in a hori- 
zontal position. 

Stadia Measurements. As already stated on page 232, the 
stadia is an attachment to the telescope used in measuring 
distances, especially over rough ground. It consists essen- 
tially of two horizontal wires fastened to small movable 
slides, and so adjusted as to include a given space, say 
one foot on a rod 100 feet distant. These wires will then 
include two feet on a rod 200 feet away, or a half-foot at a 
distance of 50 feet, and so on. Usually the instrument is so 
adjusted that the zero of the indicated distance is in front 
of the centre of the instrument ; hence, the true distance 
is the indicated distance plus the distance of this zero from 



FIELD INSTRUMENTS 235 

the centre of the instrument. This latter distance is deter- 
mined for each instrument by the maker, and noted on a card 
placed on the inside of the instrument box. It is known as 
the constant of the instrument. The readings are taken on a 
rod, specially designed for the purpose, known as the stadia 
rod. This is graduated to feet, and tenths and hundredths of 
a foot. Any ordinary leveling rod, if similarly graduated, will 
answer the same purpose. Obviously in taking stadia meas- 
urements the rod must always be held at right angles to 
the line of sight. This statement has special reference to 
measurements taken up or down hill-slopes. In this case, if 
horizontal distance is required, the measured distance must 
be multiplied by the cosine of the angle of elevation or 
depression. (Why ?) 



EXERCISE HI 

1. By means of the transit, measure the interior angles of 
the field of Example 1, Exercise II, p. 230, and compare with 
the results obtained in Example 2 of the same exercise. 

2. Lay out the entire angular magnitude about some point 
into four or more angles, and measure each one of them. What 
should the sum of them equal ? 

3. If the constant of a transit adjusted to one foot 100 feet 
away is 3.8 inches, what is the true length of a line when the 
indication on the rod is 2.35 feet ? 

4. Measure a line by the stadia, and compare with measure- 
ments taken by the chain and also by the tape. 

5. Compute the height of a tall object, as a tree or steeple, 
by first measuring its distance from some convenient point and 
measuring the angle of elevation at that point. 

6. Lay out a square field containing just one acre. 



236 SURVEYING 

SECTION VI 
THE SOLAR COMPASS 

Description and Uses. A full description of the solar com- 
pass, or Burt's solar compass, as it is often called from its 
inventor, with its principles, adjustments and uses, forms the 
subject of a considerable volume, which should be in the hands 
of the surveyor who uses this instrument. The limits of our 
space will allow only a brief reference to its principal features. 
Fig. 15 exhibits the instrument by itself ; Fig. 16, p. 239, is a 
graphical illustration of the solar apparatus as an attachment to 
the transit, the circles shown being intended to represent those 
supposed to be drawn upon the concave surface of the heavens. 
The form of the solar compass shown in Fig. 15 has the 
arrangement of its sockets and plates similar to that of the 
transit, the standards similar to those of the compass, the solar 
apparatus being placed on the upper vernier plate and taking 
the place of the needle, for which it operates as a substitute 
in the field. 

The solar compass consists mainly of three a.rcs of circles, a 
the latitude arc, by which is set off the latitude of the place, b 
the declination arc, by which is set off the declination of the 
sun, and c the hour arc, by which is set off the hour of the day. 
The arm h is movable about a point at the extremity of the 
piece containing the declination arc, there being at each end a 
solar lens having its focus on a silvered plate on the other end. 
The arc of the declination limb turns on an axis, and one or 
the other solar lens is used, according as the sun is north or 
south of the equator. Fig. 15 shows the position of the decli- 
nation arc when the sun is south ; Fig. 16, when it is north. 
The needle box is moved about its centre by a slow-motion 
screw. It contains a magnetic needle, and is furnished with a 
graduated arc about 36° in extent. 



FIELD INSTRUMENTS 



237 










Fig. 15. Burt's Solar Compass 



FIELD INSTRUMENTS 



239 




Fig. 16. Transit with Solar Attachment 

The circles shown in the cut are intended to represent in miniature circles supposed 
to be drawn upon the concave surface of the heavens. 



240 SURVEYING 

The solar compass may be used for most of the purposes of a 
compass or transit. Its most important use, however, is to run 
north and south lines, especially in laying out the public lands. 
It may be used also in determining the latitude of a place. 

To establish a True Meridian. Set off on the latitude arc 
the latitude of the place. Set off on the declination arc the 
declination of the sun, corrected for refraction. Set the 
instrument over the station ; level, and turn the sights in a 
north and south direction by the needle. The surveyor then 
turns the solar lens to the sun, and with one hand on the 
instrument and the other on the revolving arm, moves both 
from side to side until the sun's image is made to appear on 
the silvered plate, precisely between the equatorial lines. The 
line of sights then indicates the true meridian. 

The bearing of any line from the meridian may be read by 
the verniers of the horizontal limb. When a due east and 
west line is to be run, these verniers are set at 90°, and the 
sun's image is kept between the lines as before. 

Other Methods. By North Star at Culmination. The North 
Star, or Polaris, at present revolves about the north pole of 
the heavens at the distance of about 1^° ; hence, it is on the 
meridian twice in 23 h. 56 m. 4 s. (a sidereal day), once above 
the pole, called the upper culmination, and 11 h. 58 m. 2 s. 
later below the pole, called the loiver culmination. 

The time of the upper culmination of Polaris may be found 
by means of the star Mizar, the middle one of the three stars 
in the handle of the Dipper (in the constellation of the Great 
Bear). It crosses the meridian at nearly the same time as 
Polaris. Suspend a plumb line, placing the bob in a pail of 
water to lessen its vibrations. South of the plumb line, upon 
a horizontal board firmly supported, place a compass sight, or 
any upright with a small opening or slit, fastened to a board 
a few inches square. At night, when Mizar by estimation 
approaches the meridian, place the compass sight in line with 



FIELD INSTRUMENTS 



241 



Cassiopeiae 



Polaris 



Polaris and the plumb line, and move it so as to keep it in 
this line until the plumb line falls also on Mizar (Fig. 17). 
Note the time ; then (1903) 3 in. 39 s. later Polaris will be 
on the meridian. If then Polaris, the plumb line and the com- 
pass sight are brought into line, the plumb line and compass 
sight will give two points in the meridian ; or if the telescope 
of the transit is brought to bear on Polaris, 
and a light is held near to make the wires 
visible if necessary, the telescope will 
then lie in the plane of the meridian, 
which may be marked by bringing the 
telescope to a horizontal position. 

For each year subsequent to 1903 add 
21 s. to 3 m. 39 s. If the lower culmina- 
tion takes place at night, the time may be 
found in a similar manner. When Mizar 
cannot conveniently be used, 8 Cassiopeiae 
(Fig. 17) may be employed, the method 
being the same as in the case of Mizar. 
The interval, however (1903), is 4 m. 24 s. 
and the annual increase of the interval Mizar > 
about 20 s. * 

By North Star at Greatest Elongation. 
When Polaris is at its greatest apparent 
angular distance east or west of the pole, 
it is said to be at greatest elongation. It 
attains its greatest eastern elongation and 
western elongation, respectively, 5 h. 59 m. 1 s. after lower and 
upper culmination. The azimuth of a star is the angle which 
the meridian plane makes with the vertical circle passing 
through the star and the zenith of the observer. 

If now we know the time of either extreme elongation and 
also the azimuth of Polaris at an extreme elongation, we can 
from these data establish a true meridian. The latter of these 



Pole 



Great 



Bear 



() 



Fig. 17 



242 SURVEYING 

data is given for various latitudes and for years to come in 
tables, to which the surveyor is supposed to have access. To 
obtain a line in the direction of Polaris at greatest elongation, 
we may proceed as follows : A few minutes before the time 
of greatest elongation, place the compass sight in line with 
the plumb line and Polaris, keeping it in line with these until 
the star begins to recede. At this moment the sight and 
plumb line are in the required line. Or bring the telescope 
of the transit to bear on the star, and follow it keeping the 
vertical wire over the star until it begins to recede. The 
telescope will then be in the required line. In either case, 
after having the transit sighted in the direction of the line 
just found, turn it in the proper direction through an angle 
equal to the azimuth as found from the tables. 

The accompanying table * gives the Washington mean time 
of each tenth transit of Polaris (upper culmination) at the 
meridian of Washington, D.C. The last column contains the 
variation per day, to facilitate the interpolation of the time 
for any intermediate transit. 

The transit which occurs October 17 is the tenth transit 
following that which occurs on October 8. This is because 
two transits occur on October 13 ; the interval separating them 
being 23 h. 56 m. 4 s. of mean time. These two transits are 
introduced in the table for greater convenience, and as a safe- 
guard against error respecting the particular day of transits in 
that vicinity. The double lines merely call attention to the 
break thus caused in the series. 

By interpolation we may, by taking account of the longitude 
of any given station, find the local mean time of transit of 
Polaris at that station for any particular day. Thus, to find 
the Cincinnati mean time of the upper culmination of Polaris 
at Cincinnati, on May 15, 1902, we have (p. 244) : 

* Furnished by the Director of the Nautical Almanac Office, Wash- 
ington, D.C. 



FIELD INSTRUMENTS 






Day "F 


Local Mean Time 


Cha> 


the Yeah 


or Every 10th Tba 


1 Day 


1902 


. 




Jan. 1 


6M1"> 19* p.m. 


- 3" 56*. 8 


11 


6 1 51 


56.9 


21 


" 22 .. •• 


■ 


31 


4 4J 53 •• 




Feb. 10 


4 3 .4 •• 


- 


20 


- 


' ' 


Mar . 


2 44 29 •• 


56.6 


12 


- " 4 •■ 




.- 


1 25 4; •• 


' - 


Apr. 1 


12 4 . 


56.0 


11 


12 7 1 " 


' I - 


21 


11 2: 44 a.m. 




1 


10 4- 28 




11 


10 9 14 • 




21 


9 30 2 • 






8 60 1 




June 10 


8 11 41 • 




L'" 


" - 




30 


6 53 22 •• 




July 10 


14 13 •• 




20 


5 35 5 " 




30 


-.; «i 




Aug. 9 


4 16 47 • 


" 


19 


" ' • 


55 .1 


_ 


2 58 2 


" ' - 


Sept 8 


2 IS 14 •' 




18 


1 40 1 • 




28 


1 4 




Oct 8 


- - 




Oct. 13 


12 1 .'.I A.M. 




Oct 13 


11 " P.M. 




Oct. 17 


11 42 12 p.m. 


I ' - 


.- 


11 1 




N r. 6 


1" 23 :-.2 " 


56.1 


16 


.4 10 " 




. 


9 4 4 


■ - 


Dec. 6 


- -■ 21 •• 




16 


: 4;. 55 •• 


: 2 


26 


.- ■• 


- 


36 


18 


- 3 : 



244 SURVEYING 

Local mean time of transit at Washington, May 11, 1902 

= 10 h 9 m 14 s a.m. 
Longitude of Cincinnati ivest of Washington 

= + O 11 29 m 40 s = + d .021. 
May 15 d + d .021 = May 15 d .021. 
Preceding tabular date = May 11. 
Therefore, interval = 4 d 021. 

Daily variation = - 3 m 55 s .3 = - 235 s .3. 

Total change - 4.021 x (- 235 s .3) = - 15 m 46 s . 

10 h 9 m 14 s a.m. 
-15 46 
9 h 53 m 28 s a.m. 

Therefore, the required time is 9 h 53 m 28 s a.m., May 15, 1902. 



SECTION VII 
THE Y LEVEL 

Description. The essential parts of the Y level (Fig. 18) 
are, the telescope, which is of various lengths, usually about 
20 inches, and rests on supports called Y's, from their shape ; 
the spirit level, which is under the telescope and attached to 
it ; and the leveling head and tripod, which are similar to the 
same parts of the transit. 

Leveling Rod. There are several kinds of leveling rods, 
each possessing some merit peculiar to its purpose. The one 
shown in Fig. 19 is known as the Philadelphia leveling rod, 
and is the one in most common use. It is made of two pieces 
of wood, sliding upon each other, and held in position by a 
clamp. The front surface of each piece is graduated to hun- 
dredths of a foot up to 7 feet ; the back surface of the rear 
piece is figured from 7 to 13 feet, reading from the top down, 



FIELD INSTRUMENTS 



245 




Fig. 18. The Y Level 



246 



SURVEYING 



and it also has a scale by which the rod is read to half hum 
dredths of a foot as it is extended. The target 
slides along the front of the rod and is held in 
place by two springs which press upon the sides 
of the rod. It has a square opening at the centre, 
through which the division line of the rod oppo- 
site to the horizontal line of the target may be 
seen. It also carries a scale by which heights 
may be read to half hundredths of a foot. For 
heights not greater than 7 feet, the target is 
moved up or down .the front surface, the rod 
being closed and clamped; but when a greater 
height is required the target is fixed at 7 feet 
and the rear half of the rod extended to the 
required height. The rod thus becomes a self- 
reading rod 13 feet long. 

How to use Level and Rod. When the leveling 
instrument is used, the tripod should be set firm ; 
the spirit level should then be brought succes- 
sively over each opposite pair of leveling screws 
and leveled in each position, the operation being 
repeated until the bubble remains in the middle 
of the tube through an entire rotation of the tele- 
scope. Each time before taking an observation 
the instrument should be examined to see if it 
is still level. Care should be taken to bring the 
cross wires of the telescope precisely in focus and 
the object into such perfect view that the wires 
will appear to be fastened to the surface, how- 
ever the eye is moved. For very accurate work 
the instrument should be shielded from the direct 
rays of the sun. 
FlG 19 The leveling rod should be held in a truly 

vertical position, the rodman standing squarely behind it. 



FIELD INSTRUMENTS 



247 



The target is then raised or lowered at the signal of the 
leveler until its horizontal line is cut by the intersection of 
the cross wires of the telescope. The reading is done by the 
leveler or the rodman according to the kind of rod used. 

Substitutes for the Y Level. For ordinary work, the Sur- 
veyor's or Engineer's Transit is often used. 

The plumb level (Fig. 20) consists of two pieces of wood 
joined at right angles. A straight line is drawn on the upright 
perpendicular to the upper edge of the crosshead. The instru- 
ment is fastened to a support by a screw through the centre of 
the crosshead. The upper edge of the crosshead is brought 
to a level by making the line on the upright coincide with a 
plumb line. 



9 



9 



Fig. 20 



Fig. 21 



Fig. 22 



A carpenter's square can be made into a level by being 
supported by a post (Fig. 21), the top of which is split or sawed 
so as to receive the longer arm. The shorter arm is made 
vertical by a plumb line, which brings the longer arm to a 
level. 

The water level, as shown in Fig. 22, consists of two 
upright glass tubes cemented into a connecting tube of any 
material. The whole is nearly filled with water and sup- 
ported at a convenient height. The surface of the water 
in the uprights determines the level. The water should be 
colored. 



248 SURVEYING 

A level line may be obtained by sighting along the upper 
surface of the block in which an ordinary spirit level is 
mounted. 

For many purposes not requiring great accuracy, any of the 
foregoing simple instruments in connection with any graduated 
rod will be sufficient. 

EXERCISE IV 

1. Set up the level and take the readings on the leveling 
rod at two stations equally distant from the instrument. What 
does the difference of these readings indicate ? 

2. Set up the level successively at the two stations in 
Example 1, taking the readings on the leveling rod placed 
where the instrument was first. What does the difference of 
these readings indicate ? Ought this difference to be the 
same as that in Example 1? Explain. 

3. In the field of Example 1, Exercise II, p. 230, set up the 
level successively at the middle of each of the five sides, taking 
the readings on the rod each time at both adjacent stations of 
the field. Eind the difference between the sum of the hind- 
sights and the sum of the foresights. What should this 
difference equal ? 



SECTION VIII 

THE PLANE TABLE 

Description and Uses. The plane table, an approved form of 
which is shown in Eig. 23, consists mainly of a drawing 
board made of well-seasoned wood, arranged in sections to pre- 
vent warping, and supported at a convenient height by a tripod 
and leveling head, with attachments for horizontal movement. 



FIELD INSTRUMENTS 



249 




Fig. 23. The Plase Table 



250 SURVEYING 

The board is provided with rollers or clamps or both, for 
keeping the paper secure and even. The plumbing arm has 
its end brought to a point which, however placed on the 
paper, is directly above the corresponding point on the ground 
determined by the plummet. The alidade is a ruler of brass 
or steel supporting a telescope with stadia or sight standards, 
whose line of sight is in or parallel to the same vertical plane 
with the beveled edge of the ruler. A compass with two 
spirit levels serves both to level the table and, when applied 
by the edges parallel to the zero line of the compass circle, 
to determine the magnetic bearing of the lines drawn on the 
paper, or the direction of the table itself. 

After the principal lines of a survey have been determined 
and plotted, the details of the plot may be filled in by means 
of the plane tables or, when a plot only of a tract of land is 
desired and extreme accuracy is not required, this instrument 
affords the most expeditious means of obtaining it. There is 
little use for it outside of the United States Coast and Geodetic 
Survey and the United States Geological Survey. 

To orient the Table. This operation consists in placing the 
table so that the lines of the plot shall be parallel to the 
corresponding lines on the ground. 

This may be accomplished approximately by turning the 
table until the needle of the compass indicates the same bear- 
ing as at a previous station, the edge of the compass coinciding 
with the same line on the paper at both stations. 

If, however, the line connecting the station at which the 
instrument is placed with another station is already plotted, 
the table may be placed in position accurately by placing it 
over the station so that the plotted line is by estimation over 
and in the direction of the line on the ground; then making 
the edge of the ruler coincide with the plotted line, and turn- 
ing the board until the line of sight bisects the signal at the 
other end of the line on the ground. 



FIELD INSTRUMENTS 



251 



To plot any Point. Let ab on the paper represent the line 
A B on the ground ; it is required to plot c, representing C on 
the ground. 



1. By intersection. 

Place the table in position at A (Fig. 
making the fiducial edge of the 
ruler pass through a; turn the 
alidade about a until the line of 
sight bisects the signal at C, and 
draw a line along the fiducial edge 
of the ruler. Place the table in 
position at B, plumbing b over B, 
and repeat the operation just 
described. Then c is the intersec- 
tion of the two lines thus drawn. 



24), plumbing a over A, and 



< 

1 N 




























1 \ 
















X c 




v v 


































b< -a 




b 


-a 





B 



Fm. 24 



'# 



2. By resection. 

Place the table in position at A (Fig. 25), and draw a line in the direc- 
tion of C, as in the former case ; then remove the instrument to C, place 

it in position by the line drawn 
from a, make the edge of the 
ruler pass through b, and turn 
the alidade about b until B is in 
the line of sight. A line drawn 
along the edge of the ruler will 
intersect the line from a in c. 

3. By radiation. 

Place the table in position at A 
(Fig. 26), and draw a line from a 
toward C, as in the former cases. 
Measure AC, and lay off ac to 
the same scale as ab. 

To plot a Field ABCD • • • 

By radiation. 

Set up the table at any point 

P, and mark p on the paper over 

Fig. 2G P. Draw indefinite lines from p 



1 


*. N 






1 


~^ x 


1 


^ x 


1 
1 


^v 








h v -/y 


l B 





Fig. 25 



G 



X 



\ 











\ 












\0 






























b 


— a 




B 







252 



SURVEYING 



toward A, B, C, • • • Measure PA, PB, • • , and lay off pa, pb, • • • to 
suitable scale, and join a and b, b and c, c and d, • • • 



< 

















\c 
























& -a 




B 





By progression. 

Set up the table at A, and draw 
a line from a toward B. Measure 
AB, and plot ab to a suitable 
scale. Set up the table in position 
at B, and in like manner deter- 
mine and plot be ; and so on. 



Fig. 27 



By intersection. 

Plot one side as a base line. 
Plot the other corners by the method of intersection, and join these points 
in proper order by straight lines. 

By resection. 

Plot one side as a base line. Plot the other corners by the method of 
resection, and join these points in proper order by straight lines. 

The Three-Point Problem. Let A, B, C represent three field 
stations plotted as a, b, c, respectively (Fig. 28) ; it is required 




Fig. 28 



to plot d representing a fourth field station D, from which A, 
B, and C are visible. 

Place the table over D, level and orient approximately by 
the compass. Determine d by resection as follows : Make the 



• FIELD INSTRUMENTS 253 

edge of the ruler pass through a and lie in the direction a A, 
and draw a line along the edge of the ruler. In like manner, 
draw lines through b toward B and through e toward C. If 
the table is oriented perfectly, these lines meet at the required 
point d, but ordinarily they will form the triangle of error, ab, 
ac, be. In this case, through a, b, and ab ; a, e, and ac ; and 
b, c, and be, respectively, draw circles ; these circles will inter- 
sect in the required point d. For at the required point the 
sides ab, ac, be must subtend the same angle as at the points 
ab, ac, be, respectively. Hence, the required point d lies at 
the intersection of the three circles mentioned. The plane 
table may now be oriented accurately. 

The three-point problem may also be solved by fastening on the board 
a piece of tracing paper and marking the point d representing D, after 
which lines are drawn from d toward A, B, and C. The tracing paper 
is then moved until the lines thus drawn pass through a, b, c, respectively, 
when by pricking through d the point is determined on the plot below. 
This method, however, is impracticable in case the wind blows. 



CHAPTER II 
OFFICE INSTRUMENTS 

SECTION IX 
PLOTTING INSTRUMENTS 

Definitions. A map is a representation by means of points, 
lines, and conventional signs on a plane surface, as on paper 
of a surveyed portion of the earth's surface, including objects 
upon it. If only the boundary lines are drawn, the repre- 
sentation is called an outline map, or plot. The plot is a 
figure similar to the original, and the ratio of a line of the 
field to the corresponding line of the plot is called the scale. 
In surveying it is customary to designate the scale as so many 
chains to the inch. 

Principal Minor Instruments. The principal minor instru- 
ments used in plotting are a ruler, pencil, straight-line pen, 
hair-spring dividers, compasses, a right triangle of wood or 
hard rubber, a T-square, and a parallel ruler. 

The Diagonal Scale. A portion of this scale is shown in 
Fig. 29. AB is the unit. AB and A'B' are divided into ten 
equal parts, and B is joined with h, the first division point to 
the left of B'-, the first division point to the left of B is joined 
with the second to the left of B', and so on. The part of the 
horizontal line numbered 1 intercepted between BB' and Bh is 
evidently ^ of t l = T ^ of the unit ; the part of the hori- 
zontal line numbered 2 intercepted between BB' and Bh is T ^ 

of the unit, and so on. 

254 



OFFICE INSTRUMENTS 



255 



The method of using this scale is as follows : 
Let it be required to lay oft' the distance 1.43. 



A' 



h B' 



C 







\ 






I 












/ 






\ 












/ 


' 




3 











10 9 


8 


7 


G 


5 4 


3 


2 


1 


A 














B 



Fig. 29 



Place one foot of the dividers at the intersection of the horizontal line 
numbered 3 and the diagonal numbered 4, and place the other fo >t at 
the intersection of the vertical line numbered 1 (CC) and the horizontal 
line numbered 3 ; the distance between the feet of the dividers will be 
the distance required. For, measuring along the horizontal line num- 
bered 3, from CC' to BB' is 1 ; from BB' to Bh is 0.03 ; and from Bh to 
the diagonal numbered 4 is 0.4 ; and 1 -f 0.03 + 0.4 = 1.43. 

The Circular Protractor. This instrument (Fig. 30) usually 
consists of a semicircular piece of brass or german silver, with 
its arc divided into degrees and its centre marked. 

Some protractors have an arm which carries a vernier, by 
which angles may be constructed to single minutes. Still 
others embrace an entire circle and have several arms with 
verniers. 

A rectangular protractor, having the degrees marked off on 
three sides of a plane scale, is sometimes used. Often this 
form of the protractor is found on the reverse side of the 
diagonal scale, 



256 



SURVEYING 



Constructions. 1. To lay off an angle with the circular 
protractor. Place the centre over the vertex of the angle, 
and make the diameter coincide with the given side of the 
angle. Mark off the number of degrees in the given angle, 
and draw a line through this point and the vertex. 




Fig. 30 

2. To draw through a given point a line parallel to a given 
line with a right triangle and ruler. 

Make one of the sides of the triangle coincide with the 
given line, and, placing the ruler against one of the other sides, 
move the triangle along the ruler until the first side passes 
through the given point ; then draw a line along this side. 

3. To draw through a given 2>oint a line perpendicular to 
a given line with a right triangle and ruler. 

Make the hypotenuse of the right triangle coincide with 
the given line, and, placing a ruler against one of the other 
sides of the triangle, revolve the triangle about the vertex of 
the right angle as a centre until its other perpendicular side 
is against the ruler; then move the triangle along the ruler 
until the hypotenuse passes through the given point, and 
draw a line along the hypotenuse. 



OFFICE INSTRUMENTS 257 

SECTION X 
COMPUTING INSTRUMENTS 

The Planimeter. This is an instrument for measuring the 
area of any irregular field, by applying it to a plot of the 
field drawn accurately to scale. The form in most common 
use is that known as the polar planimeter. The essential 
parts are two arms, one fixed in length, the other adjustable. 
and a rolling wheel mounted on an axis parallel to the adjust- 
able arm. The outer end of the arm of fixed length is made 
fast to the plot by means of a needle point, and the free end 
of the other arm is made to trace the perimeter of the figure 
to be measured. A disk records the area in the unit for 
which the instrument is set. 

The Slide Rule. This is an instrument for effecting the 
processes of multiplication, division, involution, and evolution 
by means of logarithms. It consists of a series of scales so 
arranged that by sliding one upon the other the addition or 
subtraction of logarithms is mechanically performed. For a 
full description of this labor-saving device in its various forms, 
the student is referred to some treatise. on the subject. 



CHAPTER III 

LAND SURVEYING 

SECTION XI 

DEFINITIONS 

Land Surveying is the art of measuring, laying out, and 
dividing land, computing parts and areas from measured parts, 
and preparing a plot. An original survey includes laying out 
the boundary lines and establishing the corners. A resurvey 
is the retracing of old boundary lines and the finding of corner 
monuments, or the relocating of them when lost. 
Rules for Areas. The unit of land measure is the 

acre — 10 square chains == 4 roods 

= 160 square rods, perches, or poles. 
Areas are referred to the horizontal plane, no allowance 
being made for inequalities of surface. 

Let A, B, and C be the angles of a triangle, and a, b, and 
c the opposite sides, respectively, and let s = ^{a -f b + c). 
Area of triangle ABC = % base x altitude 
= i be sin A 

a 2 sin B sin C 
= 2 sin (B + C) 



= Vs (s — a) (s — b) (s — c). 
Area of rectangle = base x altitude. 
Area of trapezoid — \ sum of parallel sides x altitude. 

Note. Spanish American units are in use in Texas, California, and 
Mexico. In this system the vara is the unit of length, which in Texas is 

258 



LAND SURVEYING 259 

reckoned 33£ inches, in California 33 inches, in Mexico 32.9927 inches. 
The area of a square 1000 varas on a side is called a labor, and of a 
square 5000 varas on a side is called a league. 



SECTION XII 
SPECIAL METHODS OF SURVEYING, AND COMPUTING AREAS 

Triangular Fields. Measure, as may be most convenient, 
the three sides, two sides and the included angle, two angles 
and the included side, or a side and the altitude upon that 
side, and compute the area by the appropriate formula. 

Fields having More than Three Straight Sides. Divide the 
field into triangles and take the sum of the areas of the 
triangles. Or, run a diagonal and perpendiculars to it from 
the opposite vertices ; take the sum of the areas of the right 
triangles, rectangles, and trapezoids thus formed. 

A third method is as follows: Let A BCD (Fig. 31) repre- 
sent a field, and P and P' two stations within it. (They may 
be without the field.) Measure D 
PP' with great exactness. Meas- 
ure the angles between PP' and 
the lines from P and P' to the 
corners of the field. 

In the triangle P'PD, PP' and 
the angles PP'D and P'PD are 
known ; hence, PD may be found. 
In like manner, PC may be found. 
Then, in the triangle PDC, PD, 
PC, and the angle DPC are known ; hence, the area of PDC 
may be computed. In like manner, the areas of all the trian- 
gles about P or P' may be determined. 

Area ABCD = PAD + PDC + PCB + PBA; 
also, area ABCD = P'AD + P'DC + P'CB + P'BA. 




260 



SURVEYING 



Fields having Irregular Boundary Lines. Let A GBCD (Fig. 32) 
represent a field having a stream AEFGHKB as a boundary 
line. Run the line AB. From E, F, G, H, K, prominent 
points on the bank of the stream, let fall perpendiculars EE', 
FF', GG', etc., upon AB. Regarding AE, EF, etc., as straight 





Fig. 32 



Fig. 33 



lines, the portion of the field cut off by AB is divided into 
right triangles, rectangles, and trapezoids, the necessary ele- 
ments of which can be measured and the areas computed. 
The sum of these areas added to the area of A BCD gives 
the area required. If the offsets are at regular intervals, then 
the area of the part cut off by AB may be found by adding the 
offsets and multiplying by the common distance between them. 

When the irregular boundary line crosses the straight line 
that joins its extremities, as in Fig. 33, the areas of AEFH&nd 
HGB may be found separately, as in the preceding case. Then, 
the area required = ABCD + HGB — AEFH. 
. Rectangular System of Co-ordinates. Let XX' and YY' (Fig. 
34) be two fixed perpendicular lines intersecting at the point 0. 
Let the four parts into which these lines divide the plane be 
called Quadrants, as in Trigonometry, and be distinguished by 
naming them, respectively, first, second, third, and fourth 
quadrants. 

Suppose the position of a point is described by saying that 
its distance from YY', expressed in terms of some chosen unit 



J. AND SURVEYING 



261 



of length, is 3, and its distance from A" A'' is 4. Then there 
is in each quadrant one point and only one which will 
satisfy these conditions. The position of the point in each 
quadrant may be found by drawing parallels to YY' at the 
distance 3 from YY', and parallels D 

to XX' at the distance 4 from XX' ; E \p — 



then the intersections P 1} P 2 , P w 



ii 



t -v 



in 



IV 



Y 

Fig. 34 



Pt 



M 



-H 



and P 4 satisfy the given condi- 
tions. -5 

In order to determine which 
one of the four points, P x , P 2 , P 3 , 
P 4 , is meant, we adopt the rule C 
that distances measured from YY 1 
to the right are positive; to the 
left, negative. Distances measured from XX' upward are 
positive ; downward, negative. Then, the position of P x will 
be denoted by + 3, + 4 ; of P 2 , by — 3, + 4 ; of P 3 , by - 3, - 4 ; 
of P 4 , by + 3, - 4. 

The fixed lines XX' and YY' are called the Axes of Co-ordi- 
nates ; XX' is called the Axis of Abscissas, or Axis of x; YY', 
the Axis of Ordinates, or Axis of y. The intersection is 
called the Origin. 

The two distances (with signs prefixed) which determine 
the position of a point are called the Co-ordinates of the 
point ; the distance of the point from YY' is called its 
Abscissa ; and the distance from XX', its Ordinate. 

Abscissas are usually denoted by x, and ordinates by y, 
and a point is represented algebraically by simply writing 
the values of its co-ordinates within parentheses, that of the 
abscissa being always written first. 

Thus, P x (Fig. 34) is the point (3, 4), P 2 the point 
(— 3, 4), P 3 the point (— 3, — 4), and P 4 the point (3, — 4). 
In general the point whose co-ordinates are x and y is the 
point (x, y). 



262 



SURVEYING 



This system of co-ordinates may be applied to the determi- 
nation of areas in the following manner : 

Suppose the field to be ABODE (Fig. 35). Lay out the two 
axes so that the held shall lie wholly within the first quad- 
rant. Then measure the co-ordinates of each of the vertices 




and designate them as follows : for A, (x ly y t ) ; for B, (.r 2 , y 2 ) ; 
for C, (x 3 , y 3 ) ; for D, (<r 4 , y 4 ) ; for E, (x s , y 6 ). Evidently each 
of these co-ordinates is positive. Then, 

area ABCDE = area LABM + area MBCP + area PCDR 

— area NEDR — area LA EN; 

or, in terms of the co-ordinates, 

area A BCDE = j (y r + y 2 ) (x 2 — Xl ) -f % (y 2 + y z ) (x 3 - x 2 ) 

+ i (2/3 4- 2/4) 0»4 - «a) - i(2/4 + 2/5) («4 - #5) 
-i (2/5 + 2/i) (^5-^1), 

= 4- 1^1(2/5 — 2/2) + * 2 (2/1 - y*) + *3 (y 2 - 2/4) 

+ *4 (?/3 - 2/5) + ^5(2/4 - Vl) \- 

This method can be put in the form of a general rule : 
Take one-half the algebraic sum of the products obtained 
by multiplying the abscissa of each vertex by the difference 
between the ordinates of the two adjacent vertices, taken in 
the clockwise order. 



LAND SURVEYING 



26', 



EXERCISE V 



1. Required the area of a triangular field whose sides are 
13 chains, 14 chains, and 15 chains. 

2. Required the area of a triangular field if it has two 
angles 48° 30' and 71° 45', and the included side 20 chains. 

3. Required the area of a triangular field whose base is 
12.60 chains, and altitude 6.40 chains. 

4. Required the area of a triangular field which has two sides 
4.50 chains and 3.70 chains, and the included angle 60°. 

5. Required the area of a field in the form of a trapezium, 
one of whose diagonals is 9 chains, and the two perpendicu- 
lars upon this diagonal from the oppo- 
site vertices 4.50 chains and 3.25 chains. 

6. Required the area of the field 
ABCDEF (Fig. 36), if 
AE = 9.25 chains, FF' = 6.40 chains, 
BE = 13.75 chains, DD' = 7 chains, 
DB = 10 chains, CC = 4 chains, 

and A A' = 4.75 chains. fig. 36 

7. Determine the area of the field A BCD from two interior 
stations P and P', if PP' = 1.50 chains, 

PP'C = 89° 35', PP'D = 349° 45', P'PD = 165° 40', 
PP'B = 185° 30', P'PB = 3° 35', P'PC = 303° 15'. 
PP'A = 309° 15', P'PA = 113° 45', 

8. Required the area of the field 
ABCDEF (Fig. 37), if 
AF' = 4 chains, FF' = 6 chains, A 
EE' = 6.50 chains, AE' = 9 chains, 
AD = 14 chains, A C = 10 chains, 
AB' = 6.50 chains, BB' == 7 chains, 
CC = 6.75 chains. 





264 SURVEYING 

9. Kequired the area of the field AGBCD (Fig. 32, p. 260), 
if the diagonal AC = 5, BB' (the perpendicular from B to AC) 
= 1, DD' (the perpendicular from D to AC) = 1.60, EE' = 
0.25, FF' = 0.25, GG' = 0.60, HH' = 0.52, 2^'= 0.54, ,4£' = 
0.2, F'F' = 0.50, F'G' = 0.45, G'H' = 0.45, #'#' = 0.60, and 
K'B = 0.40. 

10. Kequired the area of the field AGBCD (Fig. 33, p. 260), 
if AD = 3, AC = 5, AB = 6, angle DAC = 45°, angle £,4(7" = 
30°, AE' = 0.75, 42* = 2.25, AH = 2.53, ,!<?' = 3.15, EE' = 
0.60, FF' = 0.40, and GG' = 0.75. 

11. Determine the area of the field A BCD from two exterior 
stations P and P', if PP' = 1.50 chains, 

P'PB = 41° 10', P'PD = 104° 45', PP'B = 132° 15', 
P'PA = 55° 45', PP'D = 66° 45', PP'A = 103° 0'. 
P'PC = 77° 20', PP'C= 95° 40', 

12. Find the area of the field ABCDE (Fig. 35, p. 262), if 
the co-ordinates, in chains, of the vertices taken in order are 
(1.40, 6.75), (4.60, 8.32), (9.00, 9.05), (12.15, 5.58), and 
(5.27, 1.16). 

13. Find the area of the field ABCDE (Fig. 35, p. 262), by 
measuring distances as follows : 

AL = 400 feet ; BM = 700 feet ; CP = 680 feet ; 
DR = 380 feet ; EN = 200 feet; LM = 150 feet ; 
MN = 250 feet; NP = 200 feet ; PR = 220 feet. 

14. Lay out a field of four sides, and find its area by the 
method of triangles and also by the method of rectangular 
co-ordinates. 

15. Lay out a field of six sides, and find its area by the 
method of triangles and also by the method of rectangular 
co-ordinates. 



LAND SURVEYING 265 

SECTION XIII 
GENERAL METHOD FOR FARM SURVEYS 

Definitions. A course is the bearing and length of a 
line. The latitude of a course is the distance between the 
parallels through its extremities, and is called a northing or a 
southing, as the course is northward or southward. The 
departure of a course is the distance between the meridians 
through its extremities, and is called an easting or a westing, 
as the course is eastward or westward. The meridian dis- 
tance of a point is its distance from a meridian. The double 
meridian distance of a course is double the meridian distance 
of its mid-point, and therefore equal to the sum of the meri- 
dian distances of the extremities of 
the course. 

Let AB (Fig. 38) represent a line, 
whose bearing and length are known. 
Let MN be a reference meridian ; 
and let p and p' be parallels through 
A and B, and m and m' meridians 
through the same points. Then, 
angle mAB represents the bearing FlG< 38 

of line AB. The latitude of the course AB is AE, and its 
departure EB. The meridian distance of the point B is BC 
and of A, AD. Evidently, the double meridian distance of 
the course AB is %(BC + AD). 

Again, in the triangle A EB, 

AE = ABx cos EAB, and EB = AB x sin EAB. 
Hence, latitude = distance x cos of bearing, and departure = 
distance x sin of bearing. From these formulas, the latitude 
and departure of any course may be found by means of a 
table of natural sines and cosines. They may be found also 



266 



SURVEYING 



from the Traverse Table, which is merely the tabulated results 
of the foregoing method for given courses. 

Field Notes. The field notes are kept in a book provided 
for the purpose. The page is commonly ruled in three col- 
umns, in the first of which is written the number of the 
station; in the second, the bearing of the side; and in the 
third, the length of the side. 

Field Notes 




1 


N. 20° E. 


8.06 


2 


S. 70° E. 


5.00 


3 


S. 10° E. 


10.00 


4 


N. 70° W. 


10.00 



Fig. 39 



To obtain the field notes, say of field 
ABC D (Fig. 39), place the compass 
at A, the first station, and take the 
bearing of AB (p. 220) ; suppose it to 
be K 20° E. Write the result in 
the second column of the field notes 
opposite the number of the station. 
Measure AB = 8.66 chains, and write 
the result in the third column of the 
field notes. Place the compass at B, 
and, after testing the bearing of AB (p.* 221), take the bear- 
ing of BC, measure BC, and write the results in the field 
notes; and so continue until the bearing and length of each 
side have been recorded. 

Computation of the Area. The survey may begin at any 
corner of the field ; but, for computing the area, the field notes 
should be arranged so that the most eastern or the most western 
station shall stand first. For the sake of uniformity, we shall 
always begin with the most western station and keep the field 
on the right in passing around it. 



LAND SURVEYING 



267 



The field notes occupy the first three of the eleven columns in the tablet 
below. Columns IV, V, VI, and VII contain the latitudes and departures 
corresponding to the sides, taken from the Traverse Table. The line 
represented by each number is indicated immediately above that number. 
Column VIII contains the meridian distances of the points J5, C, B, and 
A, taken in order. Column IX contains the double meridian distances 



I 


II 


Ill 


IV 


v 


YI 


VII 


VIII 


IX 


X XI 


Side 


Bearing 


DlST. 


N. 


S. 


E. 


W. 


M.D. 


D.M.D. 


N.A. 


S.A. 


AB 


N.20°E. 


8.66 


AB' 
8.14 




BB' 
2.96 




BB' 
2.96 


BB' 

2.96 


2 ABB' 
24.0944 




BC 


S. 70° E. 


5.00 




irr 

1.71 


C"C 
4.70 




CC 
7.66 


BB'+CC 
10.62 




2 C'CBB' 
18.1G02 


CD 


S. 10° E. 


10.00 




CD' 

9.85 


D"D 

1.74 




DD' 

9.40 


CC + DD' 

17.06 


2 f>J)Cr 
. . . . 168.0410 


DA 


N.70°W. 


10.00 


DA 

3.42 






DD' 
9.40 





DD' 

9.40 


2 ADD 1 

32.1480 .... 






33.66 


11.56 


11.56 


9.40 


9.40 






56.2424' 186.2012 



(186.2012 sq. ch. - 56.2424 sq. ch.) -*- 2 = 64.98 sq. ch. = 6.50 acres. 

of the courses. Their composition is indicated by the letters immedi- 
ately above the numbers. Column X contains the products of the double 
meridian distances by the northings in the same line. The first number, 
24.0944 = 2.96 x 8.14 = BB' x AB' = twice area of triangle ABB' ; 
32.1480 = 9.40 x 3.42 - BB' x AB' = twice area of triangle ABB'. 
Column XI contains the products of the double meridian distances by 
the southings in the same line. The first number, 

18.1602 = 10.62 x 1.71 = {BB' + CC) x B'C 

= twice area of trapezoid C'CBB' ; 
168.0410 - 17.06 x 9.85 = {CC + BB') x B'C 

= twice area of trapezoid B'BCC. 
The sum of the north areas in column X 

= 56.2424 = 2 {ABB' + ABB'). 
The sum of the south areas in column XI 

= 186.2012 = 2 {C'CBB' + B'BCC). 
But {C'CBB' + B'BCC') - {ABB' + ABB') = ABCB. 

Hence, 2 {C'CBB' + B'BCC) - 2 {ABB' + ABB') = 2 ABCB ; 
that is, 186.2012 - 56 2424 = 129.9588 = 2 ABCB. 

Hence, area ABCB = \ of 129.9588 sq. ch. = 64.98 sq. ch. = 6.50 A. 



268 SURVEYING 

Balancing the Work. In the survey, we pass entirely around 
the field ; hence, we move just as far north as south. There- 
fore, the sum of the northings should equal the sum of the 
southings. In like manner, the sum of the eastings should 
equal the sum of the westings. In this way the accuracy of 
the field work may be tested. 

In the example on page 267 the sum of the northings is 
equal to the sum of the southings, being 11.56 in each case ; and 
the sum of the eastings is equal to the sum of the westings, 
being 9.40 in each case. Hence, the work balances. 

In actual practice the work seldom balances. When it 
does not balance, corrections are generally applied to the 
latitudes and departures by the following rules : 

1. The perimeter of a field is to any one side as the total 
error in latitude is to the correction required. 

2. The perimeter of a field is to any one side as the total 
error in departure is to the correction required. 

Example. The perimeter of a field measured 306.62 chains 
and one side 72.47 chains, with a total error of 22 links in 
latitude and of 18 links in departure. 

Then 306.62 : 72.47 = 22 links : x = 18 links : y. 

Whence x — 5 links and y = 4 links. 

Hence the correction in latitude applied to the given side is 
0.05 chains, and the correction in departure is 0.04 chains. 

If special difficulty was found in taking a particular bear- 
ing, or in measuring a particular line, the corrections should 
be applied to the corresponding latitudes and departures. 

The amount of error allowable varies in the practice of dif- 
ferent surveyors, and according to the nature of the ground. 
An error of 1 link in 8 chains would not be considered too great 
on smooth, level ground ; while on rough ground an error of 
1 link in 3 chains might be allowed. If the error is consider- 
able, the field measurements should be repeated. 



LAND SURVEYING 



269 



As another example let it be required to find the area of 
field ABCDEF from the following 



Field Notes 



1 


N. 73° 30' W. 


5.00 


2 


S. 16° 30' W. 


5.00 


3 


N. 28° 30' W. 


7.07 


4 


N. 20° 00' E. 


11.18 


5 


S. 43° 30' E. 


5.00 


6 


S. 13° 30' E. 


10.00 



Side 


Bearing 


Dlst. 


N. 


S. 


E. 


W. 


M.D. 


D.M.D. 


N.A. 


S A. 


AB 
BC 
CD 
DE 
EF 
FA 


N.20°00'E. 

S.43°30'E. 

S.13 8 30'E. 

N. 73°30'W. 

S16°30'W. 

N.28°30'W. 


11.18 
5.00 

10.00 
5.00 
5.00 
7.07 


10.51 

1.42 
6.21 


3.63 

9.72 

4.79 


3.82 
3.44 
2.33 


4.79 
4.80 
1.42 

3.37 


BB 

3.82 
C'C 
7.26 
D'D 
9.59 
EE 
4.79 
FF 
3.37 

0.00 


BB 

3.82 

B'B+C'C 

11.08 
CC+D'D 

16.85 
D'D + EE 

14.38 
EE + FF 

8.16 

FF 

3.37 


2 ABB' 
40.1482 

2D DEE' 
20.4196 

2AFF 

20.9277 


2 (■(/:/: 

40.2204 

2 /)/)(( 
163.782.; 

2FFEF/ 
39.0864 






43.25 


18.14 


18.14 


9.59 


9.58 
9.59 






81.4955 


243.0888 



43.25:5 = 0.01 : x. 



Area = 8.08 acres. 



The first station in the field notes is D, but we rearrange the numbers 
in the tablet so that A stands first. The northings and southings balance, 
but the eastings exceed the westings by 1 link. We apply the correction 
to the westing 4.79 (the distance DE being in doubt), making it 4.80, and 
write the correction. In practice, the corrected numbers are written in 
red ink, and often all the latitudes and departures are rewritten in four 
additional columns, headed, respectively, N', S', E', W." 

Supplying Omissions. If for any reason the bearing and the 
length of any side do not appear in the field notes, the lati- 
tude and departure of this side may be found in the following 
manner : 



270 



SURVEYING 



Find the latitudes and departures of the other sides as 
usual. The difference between the northings and southings 
gives the northing or southing of the unknown side, and the 
difference between the eastings and westings gives the easting 
or westing of the unknown side. 

If the length and the bearing of the unknown side are de- 
sired, they may be found by solving the right triangle, whose 
sides are the latitude and departure found by the method just 
explained, and whose hypotenuse is the length required. 

Obstructions. If the end of a line is not visible from 
its beginning, or if the line is inaccessible, its length and 
bearing may be found as follows : 

By means of a random line (p. 216). 

When it is impossible to run a random line, which is fre- 
quently the case on account of the extent of the obstruction, 
the following method may be used : 



N 



E 




.-~-"0 



Let AB (Fig. 40) represent an inaccessible line 
whose extremities A and B only are known, and B 
invisible from A. 

Set flagstaffs at convenient points, C and D. 
Find the bearings and lengths of AC, CD, and DB, 
and then proceed to find the latitude and departure 
of AB. 



Fig. 40 



Example. 

lowing notes (see Fig. 40) : 



Suppose that we have the fol- 



Side 


Bearing 


DlST. 


N. 


s. 


E. 


w. 


AC ' 

CD 

DB 


S. 45° E. 

E. 

N. 30° E. 


3.00 
3.50 
4.83 


4.18 


2.12 


2.12 
3.50 
2.42 






4.18 


2.12 


8.04 






LAND SURVEYING 



271 



The northing of A B is AE = 2.06, and the easting, EB = 8.04. These 
numbers may be entered in the tablet in the columns N. and E., opposite 
the side AB. 

If the bearing and length of AB are required, 



BE _ 8.04 
^~2.00 
75° 38'. 



= 3.903. 



tan BAE = 

Hence, the angle BAE = 

Also, AB = ^AE 2 + BE 2 = V8.01 2 + 2.06 2 = 8.30. 

Therefore, the bearing and length of AB are N. 75° 38' E. and 8.30. 



To make a Plot. A plot or map may be drawn to any 
desired scale. If a line 1 inch in length in the plot represents 
a line 1 chain in length, the plot is 
said to be drawn to a scale of 1 chain 
to an inch. In this case (Fig. 41) the 
plot is drawn to a scale of 8 chains 
to an inch. 

Draw the line NAS to represent 
the meridian, and lay off the first 
northing AB' = 8.14. Through B 
draw an indefinite line perpendicular 
to NS and lay off B'B, the first easting, 
= 2.96. Draw AB ; then the line A B 
represents the first side of the field. 
Through B draw BC" perpendicular 
to BB', and make BC" = 1.71, the first 
southing. Through C" draw C"C 
perpendicular to BC", and equal to 
4.70, the second easting. Draw BC. 
The line BC represents the second 
side of the field. Proceed in like 
manner until the field is completely 
represented. The extremity of the 
last line F'A, measured from F', should fall at A. 
test of the accuracy of the plot. 




Fig. 41 



This is a 



272 



SURVEYING 



By drawing AC, AE, and EC, the hexagonal figure 
ABCDEFA is divided into triangles, the bases and altitudes 
of which may be measured and the area computed approxi- 
mately. 

Another method is as follows : 
Draw MN (Fig. 42) to represent a 
meridian. Let the point A in this 
line be taken as the first station in 
the rearranged field notes of page 
269. With the circular protractor 
mark off each of the bearings as b, c, 
d, e, f, and a. Draw AB to scale 
through b. With triangle and ruler 
(p. 256) or with parallel ruler draw to 
scale BC parallel to Ac ; and so on. 

After some practice, still other 
methods will be suggested, but the 
methods given are among the best. 




EXERCISE VI 

Find the areas of the following and make a plot of each. 
In 3 and 7, detours were made on account of obstructions 
(p. 270). The notes of the detours are written in braces. 

12 3 



St a, 



Bearings 



S. 75° E. 
S. 15° E. 
S. 75° W. 

N.45°E. 
N. 45° W. 



DlST. 



6.00 
4.00 
6.93 
5.00 
5.19* 



St a. 


Bearings 


DlST. 


1 


N. 45° E. 


10.00 


2 


S. 75° E. 


11.55 


3 


S. 15° W. 


18.21 


4 


N. 45° W. 


19.11 



Sta. 



Bearings 



S. 2°15'E. 
N. 51°45 / W. 
S. 85°00'W, 
S. 55°10 , W 
N. 3°45'E. 

s. 66°45'e. 

N. 15°00'E. 
S. 82°45 / E. 



DlST. 



9.68 
2.39 
6.47 
1.62 
6.39 
1.70 
4.98 
6.03 



LAND SURVEYING 



273 



St a. 
1 


Bearings 


DlST. 


N. 5°30'W. 


6.08 


2 


S. 82°30'W. 


6.51 


3 


S. 3°00'E. 


5.33 


4 


E. 


6.72 



Sta. Bearings Dint 



1 N. 6°15'W. 

2 S. 81°50'W. 

3 S. 5°00'E. 

4 iN.88°30'E. 



6.31 
4.06 
5.86 
4.12 



Sta. 


Bearings 


DlST. 


1 


N. 20<WE. 


4.62* 


2 


N. 73°00'E. 


4.16* 


3 


S. 45°15'E. 


6.18* 


4 


S. 38°30'W. 


8.00 


6 


Wanting 


Wanting 



•STA. 


Bearings 


DlST. 


,r 


s. 8i°20'w. 


4.28 


M 


N. 76°30'W. 


2.67 


2 


N. 5°00'E. 


8.68 


3 


S. 87°30'E. 


5.54 


r 


S. 7°00'E. 


1.79 


< 


S. 27°00'E. 


1.94 


S. 10°30'E. 


5.35 


i 


N. 76°45'W. 


1.70 



STA. 


Bearings 


DlST. 


1 

2 

j 

6 


N. 89°45'E. 
S. 7°00'W. 
S. 28°00'E. 
S. 0°45'E. 
N. 84°45'W. 
N. 2°30'\V. 


4.94 
2.30 
1.52 
2.57 
5.11 
5.79 



9. An Ohio farm is bounded and described as follows : 
Beginning at the southwest corner of lot No. 13, thence N. 1^° 
E. 132 rods and 23 links to a stake in the west boundary 
line of said lot; thence S. 89° E. 32 rods and 15 T 4 5 links to 
a stake; thence N. 1\° E. 29 rods and 15 links to a stake 
in the north boundary line of said lot ; thence S. 89° E. 61 
rods and 18^ links to a stake ; thence S. 32^° W. 54 rods 
to a stake; thence S. 35J° E. 22 rods and 4 links to a 
stake ; thence S. 48° E. 33 rods and 2 links to a stake ; 
thence S. 7£° W. 76 rods and 20 links to a stake in the south 
boundary line of said lot ; thence N. 89° AY. 96 rods and 10 
links to the place of beginning. Containing 85.87 acres, more 
or less. 

Verify the area given and plot the farm. 



274 SURVEYING 

Modification of the Latitude and Departure Method. The area 
of a field may be found by a modification of the latitude and 
departure method, if its sides and interior angles are known. 

Let A, B, C, D represent the interior angles of the field 
A BCD (Fig. 43). Let the side AB determine the direction of 
reference. The bearing of AB, with reference to AB, is 0°. 
The bearing of BC, with reference to AB, is the angle 
h = 180° — B. The bearing of CD, with reference to AB, is 
tlje angle c = C — b. The bearing of DA, with reference 

to AB, is the angle d = A. 

The area may now be com- 
puted by the latitude and depar- 
ture method, regarding AB as the 
meridian. 

In practice, the exterior angles, 

when acute, are usually measured. 

As the interior angles may be 

measured with considerable accu- 

FlG ' 43 racy by the transit, the latitudes 

and departures should be obtained by using a table of natural 

sines and cosines. 

EXERCISE VII 

1. Find the area of the field ABCD, in which the angle 
,4 = 120°, B = 60°, C = 150°, and D = 30°; and the side 
AB = 4 chains, BC = 4 chains, CD = 6.928 chains, and DA = 
8 chains. 

Keep three decimal places, and use the Traverse Table. 

2. Find the area of the farm ABCDE, in which the angle 
A = 106° 19', B = 99° 40', C = 1 20° 20', D = 86° 8', and E = 
127° 33'; and the side AB = 79.86 rods, 50 = 121.13 rods, 
CD — 90 rods, DE = 100.65 rods, and EA = 100 rods. 

Use the table of natural sines and cosines, keeping two decimal places 
in the results. 




LAND SURVEYING 275 

General Remarks on determining Areas. Operations depend- 
ing upon the reading of the magnetic needle must lack 
accuracy. Hence, when great accuracy is required (which is 
seldom the case in land surveying) the method of pp. 266-269 
cannot be employed. 

The best results are obtained by the methods explained on 
pp. 259-262 and 274, the horizontal angles being measured with 
the transit, and great care exercised in measuring the lines. 



SECTION XIV 
LOCATION SURVEYS 

Definition. In surveying proper we measure lines and 
angles as we find them, while in location surveys we mark them 
out on the ground where they are required to be in order to 
inclose a given area, or conform to a specified shape, or meet 
some other given condition. Laying out, parting off, and 
dividing up land are included in this class of surveys. The 
surveyor must, for the most part, depend on his general knowl- 
edge of Geometry and Trigonometry, and his own ingenuity, 
for the solutions of problems that arise in location surveys. 

Illustrative Problems. Problem 1. To divide a trian- 
gular field into two parts having a given 
ratio, by a line through a given vertex. 

Let ABC (Fig. 44) be the triangle, and A the 
given vertex. 

BT) 

Divide BC at D, so that equals the given 

ratio, and draw AD. ABD and ADC are the parts 
required; for 

ABD: ADC = BD:DC. 




Problem 2. To cut off from a triangular field a given 
area, by a line parallel to the base. 



276 



SURVEYING 




Let ABC (Fig. 45) be the triangle, and 
let DE be the division line required. 

Then ABC : ABE = AB 2 : AD 2 . 
•• VaBC : Va~DE = AB : AD. 



AD = AB 



VI 



ADE 



BC 



iC Problem 3. To cut off from a 
fig. 45 triangular field a given fraction of 

the field, by a line from a given point in a side. 

Let ABC (Fig. 46) be the triangle, and P the point from which the 
line PD is to be located so as to cut off, say, one-third the area of the 
triangle. 

AD = ABx AC + SAP. 

For ABC.APD = AB x AC:AP x AD = 3:1. 




Fig. 46 




Problem 4. To divide any field into two parts having a 
given ratio, by a line through a given point in the perimeter. 

Let ABCDE (Fig. 47) represent the field, P the given point, and PQ 
the required division line. 

The areas of the whole field and of the required parts having been 
determined, run the line PD from P to a corner D, dividing the field, 
approximately, as required. Determine the area PBCD. 

The triangle PDQ represents the part which must be added to PBCD 
to make the required division. 



LAND SURVEYING 



Hence, 



Note. DQ 



Area PDQ 



i x PD x DQ x sin PDQ. 

2 x area PDQ 
PD x sin PDQ ' 



2 x area PDQ 



This perpendicular from 



perpendicular from P on Di? 
P on DE may be run and measured directly. 

Problem 5. To divide a field into a given number of parts, 
so that access to a pond of water is 
given to each. 

Let ABODE (Fig. 48) represent the field, 
and P the pond. Let it be required to divide 
the field into four parts. Find the area of 
the field and of each part. 

Let AP be one division line. Run PE, 
and find the area APE. Take the differ- 
ence between APE and the area of one 
of the required parts ; this gives the area of 
the triangle PQE, from which QE may be 
found, as in Problem 4. Draw PQ ; PAQ is 
one of the required parts. In like manner, 
PQR and PAS are determined; whence, 
PSR must be the fourth part required. 




EXERCISE Vffl 

1. From the square ABCD, containing 6 acres 1 rood 24 
perches, part off 3 acres by a line EF parallel to AB. 

2. From the rectangle ABCD, containing 8 acres 1 rood 
24 perches, part off 2 acres 1 rood 32 perches by a line EF 
parallel to AD which is equal to 7 chains. Then, from the 
remainder of the rectangle, part off 2 acres 3 roods 25 perches, 
by a line GH parallel to EB. 

3. Part off 6 acres 3 roods 12 perches from a rectangle ABCD, 
containing 15 acres, by a line EF parallel to AB\ AD being 
10 chains. 



278 SURVEYING 

4. From a square ABCD, whose side is 9 chains, part off a 
triangle which shall contain 2 acres 1 rood 36 perches, by a 
line BE drawn from B to the side AD. 

5. From ABCD, representing the rectangle, whose length is 
12.65 chains, and breadth 7.58 chains, part off a trapezoid 
which shall contain 7 acres 3 roods 24 perches, by a line BE 
drawn from B to the side DC. 

6. In the triangle ABC, AB = 12 chains, AC = 10 chains, 
and BC = 8 chains ; part off a trapezoid of 1 acre 2 roods 16 
perches, by the line DE parallel to AB. 

7. In the triangle ABC, AB = 26 chains, AC = 20 chains, 
and BC = 16 chains ; part off a trapezoid of 6 acres 1 rood 24 
perches, by the line DE parallel to AB. 

8. It is required to divide the triangular field ABC among 
three persons whose claims are as the numbers 2, 3, and 5, so 
that they may all have the use of a watering place at C ; AB 
= 10 chains, AC — 6.85 chains, and CB = 6.10 chains. 

9. Divide the five-sided field ABCHE among three persons, 
X, Y, and Z, in proportion to their claims, X paying $500, Y 
paying $750, and Z paying $1000, so that each may have the 
use of an interior pond at P, the quality of the land being 
equal throughout. Given AB = 8.64 chains, BC = 8.27 chains, 
CH = 8.06 chains, HE = 6.82 chains, and EA = 9.90 chains. 
The perpendicular PD upon AB = 5.60 chains, PD' upon BC 
= 6.08 chains, PD" upon CH = 4.80 chains, PD'" upon HE 
— 5.44 chains, and PD"" upon EA = 5.40 chains. Assume 
PH as the divisional fence between the shares of X and Z, it 
is required to determine the position of the fences PM and PN 
between the shares of X and Y and between the shares of Y 
and Z. 

10. Divide the triangular field ABC, whose sides AB, AC, 
and BC are 15, 12, and 10 chains, respectively, into three equal 



LAND SURVEYING 279 

parts, by fences EG and DF parallel to BC, without finding 
the area of the field. 

11. Divide the triangular field ABC, whose sides AB, BC, 
and AC are 22, 17, and 15 chains, respectively, among three 
persons, A, P>, and C, by fences parallel to the base AB, so that 
A may have 3 acres above the line AB, B 4 acres above A's 
share, and C the remainder. 



SECTION XV 
LAYING OUT THE PUBLIC LANDS 

Reference Lines. The public lands north of the Ohio Kiver 
and west of the Mississippi are generally laid out in accord- 
ance with what is known as the rectangular system of 
surveying. First, an initial point is selected with great care, 
and then astronomically established. Through this point a 
'principal meridian, or true north and south line, is run by 
means of the solar compass, or the transit with observations 
on Polaris ; and also an east and west line, called a base line. 
Crossing the principal meridian at intervals of 24 miles, both 
north and south of the initial point, are run other east 
and west lines, called standard parallels, or correction lines. 
Northward from the base line and from each of the standard 
parallels, at intervals of 24 miles, both ways from the princi- 
pal meridian, are run true north and south lines, called guide 
meridians. Thus, the land is divided into blocks approxi- 
mately 24 miles square. Six principal meridians have been 
established, in addition to which and connected with which 
there are twenty or more independent meridians in the 
western states and territories. 

Division from Reference Lines; Townships. Within each 
block parallels to the base line, or to a standard parallel, are 
run at intervals of 6 miles. These are called township lines. 



280 



SURVEYING 



At the same intervals are also run north and south lines, 
called range lines. Thus, the tract would be divided into 
townships exactly 6 miles square if it were not for the con- 
vergence of the meridians on account of the curvature of the 
earth. An east and west series of townships is called a tier, 
and a north and south series is called a range. A township 
is designated by giving the number of the tier north or south 
of the base line and the number of the range east or west of 

N 



dW 






d 


l\k l 


1 i 


\ M 










t" 








p" 


\ B 






t' 








V' 










t 




A 




P 


w 1 










r 


r' 


r« 




a ' 











( 


-L' 






























D- — ' 


C 





























s 

Fig. 49 

the principal meridian. Thus, T. 3 N., R. 2 W., read town- 
ship three north, range two west, means that the township is 
in the third tier north of the base line, and in the second tier 
west of the principal meridian. 

Let NS (Fig. 49) represent a principal meridian ; WE a base 
line; DL and D'L' standard parallels; GM and G'M' guide 



LAND SURVEYING 



281 



meridians ; rl, r'l', . . . , range lines ; tp, t'p', . . . , township 
lines. If Or is taken as 6 miles, then O'l will be less than 
6 miles. O'k being equal to 6 miles and O'l being less, 
it will be observed that there will be offsets on the base line 
and on standard parallels at intervals of 6 miles. 

Township A would be designated thus : T. 2 N., R. 3 E. 
How would townships B and C be designated ? 

Subdivision of Townships. The townships are divided into 
sections approximately 1 mile square, and the sections are 
divided into quarter sections. The 
township, section, and quarter-section 
corners are permanently marked. The 
sections are numbered, beginning at 
the northeast corner, as in Fig. 50, 
which represents a township divided 
into sections. The quarter sections are 
designated, according to their position, 
as N.E., N.W, S.E., and S.W. Section 
lines are surveyed in such an order as 

to throw the errors on the northwest quarter sections, which 
are carefully measured and their areas calculated. 

Meander Lines. If in running a line a navigable stream 
or a lake more than 1 mile in length is encountered, it is 
meandered by marking the intersection of the line with the 
bank and running lines from this point along the bank to 
prominent points which are marked, and the lengths and 
bearings of the connecting lines recorded. 

Manual. For detail of methods, see the "Manual of Sur- 
veying Instructions," issued by the Commissioner of the 
General Land Office, at Washington, D.C., for the use of 
Surveyors-General. 



6 


5 


4 


3 


2 


1 


7 


8 


9 


10 


It 


12 


18 


17 


16 


15 


14 


13 


19 


20 


n 


22 


23 


24 


30 


29 


28 


27 


26 


25 


31 


32 


33 


34 


35 


36 



Fig. 50 



CHAPTER IV 

TRIANGULATION 

SECTION XVI 

DEFINITIONS 

The third method of surveying explained on page 259 is 
an example of triangulation on a small scale. The simple 
principle there involved is elaborately worked out in hydro- 
graphic or topographic surveys, or in the measurement of 
terrestrial arcs, as in the " Transcontinental Triangulation 
and American Arc of the Parallel," recently completed by 
the United States Coast and Geodetic Survey. 

Let F (Fig. 51) represent a point whose position with 
reference to the base line AB is required. Connect AB with 
F by the series of triangles 
ABC, ACD, ADE, and DEF, 
so that a signal at C is visible 
from A and B, a signal at D 
visible from A and C, a signal 
at E visible from A and D, 
and a signal at invisible from 
D and E. In the triangle ABC, 
the side AB is known, and the angles at A and B may be 
measured; hence, AC may be computed. In the triangle 
ACD, AC is known, and the angles at A and C may be 
measured; hence, AD may be computed. In like manner, 
BE and EF or DF may be determined. DF, or some suitable 
line connected with DF, may be measured, and this result 

282 




TRIANGULATION 283 

compared with the computed value to test the accuracy of 
the field measurement. This net or chain of triangles enables 
us to determine the relative position of all the points with 
respect to each other. If the point A is, furthermore, 
astronomically located, and the azimuth of line AB is known, 
then we have sufficient data also to determine the absolute 
geographical position of each of the points. 

Classification. Three orders of triangulation are recognized, 
viz.: primary, in which the sides are from 20 to 190 miles in 
length; secondary, in which the sides are from 5 to 40 miles 
in length, and which connect the primary with the tertiary ; 
tertiary, in which the sides are seldom over 5 miles in length, 
and which bring the survey down to such dimensions as to 
admit of the minor details being filled in by the compass and 
plane table. 

Measurement of Base Lines. Base lines should be measured 
with a degree of accuracy corresponding to their importance. 
Suitable ground must be selected and cleared of all obstruc- 
tions. Each extremity of the line may be marked by cross 
lines on the head of a copper tack driven into a stub which 
is sunk to the surface of the ground. Poles are set up in 
line about half a mile apart, the alignment being controlled 
by a transit or theodolite placed over one end of the line. 
The preliminary measurement may be made with an iron 
wire about one-eighth of an inch in diameter and 60 meters 
in length, or with a steel chain of the same length. 

The final measurement is made with the tape line, or with 
bars 6 meters long, which are supported upon trestles when in 
use. These bars are placed end to end, and brought to a 
horizontal position, if this can be quickly accomplished; if 
not, the angle of inclination is taken by a sector, or a vertical 
offset is measured with the aid of a transit, so that the exact 
horizontal distance can be computed. A thermometer is 
attached to each bar, so that the temperature of the bar may 



284 SURVEYING 

be noted and a correction for temperature applied. Some- 
times the bars are laid in melting ice, in which case accuracy 
to at least one five-millionth part of the length measured is 
attainable. 

Measurement of Angles. Angles are measured by means of 
the transit with much greater accuracy than with the com- 
pass, since the reading of the plates of the transit is taken to 
minutes, and by means of microscopes to seconds, while the 
reading of the needle of the compass is to quarter or half- 
quarter degrees. 

In order to eliminate errors of observation and of adjust- 
ment, and errors arising from imperfect graduation of the 
circles, a large number of readings is made and their mean 
taken. Two methods are in .use, viz., repetition and series. 

The method of repetition consists essentially in taking as 
many readings of an angle as is desired, the reading in each 
case after the first being from the index of the next preceding 
reading, and then taking the mean. 

The method of series is the one generally used when several 
angles about the same point are to be measured. It con- 
sists essentially in taking the readings successively on each 
station, then reversing the telescope and repeating the obser- 
vations in the reverse order, which completes a series. This 
process is repeated a number of times, each series beginning 
with a different index. Then the mean of the different series 
is found. 

On account of the curvature of the earth, the sum of the 
three angles of a triangle upon its surface exceeds 180°. 
This spherical excess, as it is called, becomes appreciable only 
when the sides of the triangle are about 5 miles in length. 
To determine the angles of the rectilinear triangle having the 
same vertices, one-third of the spherical excess is generally 
deducted from each spherical angle. 



CHAPTER V 

LEVELING 

SECTION XVII 

DEFINITIONS 

A level surface is a surface parallel with the surface of 
still water, and is, therefore, slightly curved owing to the 
spheroidal shape of the earth. A level line is a line in a 
level surface. The line of apparent level of a place is a 
tangent to the level line at that place. Hence, the line of 
apparent level is perpendicular to the plumb line. 

Leveling is the process of finding the difference of level of 
two places, or the distance of one place above or below a level 
line through another place. 

Corrections for Curvature and Refraction. In ordinary leveling 
no distinction is made between true and apparent levels. In 
precise leveling the difference between the two is measured, 
i.e., correction is made for curvature of the earth. There is 
sometimes also a correction made for refraction of light. 

Let t (Fig. 52) represent the line of apparent level of the 
place P, a the level line, d the diameter 
of the earth ; then c represents the cor- 
rection for curvature. To compute the 
correction for curvature : 



t 2 = c(c + d). (Geometry, § 381.) 

t 2 a 2 

Therefore, c = - = — > approxi- 

c + d d r 

mately, since c is very small compared 

with d, and t — a, very nearly. F! G 52 

285 




286 



SURVEYING 



Since d is constant (= 7920 miles, nearly), the correction 
for curvature varies as the square of the distance. 

Example. What is the correction for curvature for 1 mile ? 
By substituting in the formula deduced above, 

miles = 8 inches, nearly. 



d 7920 

Hence, the correction for curvature for any distance may 
be found in inches, approximately, by multiplying 8 by the 
square of the distance expressed in miles. 

If correction for refraction is also made, it is customary to 
diminish the above by about one-sixth of itself ; or, c = f of 8 a 2 . 



SECTION XVIII 
DIFFERENTIAL LEVELING 

Single Setting of Instrument. To find the difference of 
level between two places when both are visible from some 
intermediate point, and the difference of level does not 
exceed 13 feet, only one setting of the level will usually be 
necessary. 

Let A and B (Fig. 53) represent the two places. Set the 
Y level at a station equally distant, or nearly so, from A and 



A' 




Fig. 53 



B, but not necessarily on the line AB. After leveling the 
instrument, bring the telescope to bear upon the rod (p. 246), 
and by signal direct the rodman to move the target until its 
horizontal line is in the line of apparent level of the telescope. 



LEVELING 



28'* 



Let the rodraan now record the height A A' of. the target. In 
like manner find BB'. The difference between A A' and BB' 
is the difference of level required. If the instrument is 
equally distant from A and B, or nearly so, the curvature and 
the refraction on the two sides of the instrument balance, and 
no correction for curvature or refraction is necessary. 

Several Settings of Instrument. When both places are not 
visible from the same place, or when the difference of level 
between them is considerable, two or more settings of the 
level may be necessary. 

Let A and D (Fig. 54) represent the two places. Place the 
level midway between A and some intermediate station B. 




Fig. 54 



Find A A 1 and BB 1 , as in the preceding case, and record the 
former as a backsight and the latter as a foresight. Select 
another intermediate station C, and in like manner find the 
backsight BB" and the foresight CC ; and so continue until 
the place D is reached. 

The difference between the sum, of the foresights and the sum 
of the backsights will be the difference of level required. 

Since, BB' + CC + DD' - (A A' + BB" + CC") 
== BB' - BB" + CC - CC" + DD' - A A' 
= B'B" + CC" + D'D - A A' = A' A" - A A' = A A". 



288 SURVEYING 

SECTION XIX 
PROFILE LEVELING 

Definitions. The intersection of a vertical plane with the 
surface of the earth is called a section, or profile. The term 
" profile," however, usually designates the plot, or representa- 
tion of the section on paper. 

Profile leveling is leveling to obtain the data necessary for 
making a profile or plot of any required section. ' 

A profile is made for the purpose of exhibiting in a single 
view the inequalities of the surface of the ground for great 
distances along the line of some proposed improvement, such 
as a railroad, canal, or ditch, thus facilitating the establishment 
of the proper grades. 

The data necessary for making a profile of any required 
section are the heights of its different points above some 
assumed horizontal plane, called the datum plane, together 
with their horizontal distances apart or their distances from 
the beginning of the section. 

The position of the datum plane is fixed with reference to 
some permanent object near the beginning of the section, 
called a bench mark, and in order to avoid negative heights 
is assumed at such a distance below this mark that all the 
points of the section shall be above it. 

The heights of the different points of the section above 
the datum plane are determined by means of the level and 
leveling rod; and the horizontal length of the section is 
measured with an engineer's chain or tape, and divided into 
equal parts, usually 100 feet in length, called stations, marked 
by stakes numbered 0, 1, 2, 3, and so on. 

Where the ground is very irregular, it may be necessary, 
besides taking sights at the regular stakes, to take occasional 
sights at points between them. If, for instance, at a point 



LEVELING 



289 



40 feet in advance of stake 3 (Fig. 55) there is a sudden 
rise or fall in the surface, the height of this point would be 
determined and recorded as at stake 3.40. 

The readings of the rod are ordinarily taken to the nearest 
tenth of a foot, except on bench marks and points called 
turning points, where they are taken to thousandths of a foot. 

A turning point is a point on which the last sight is taken 
just before changing the position of the level, and the first 
sight from the new position of the instrument. A turning 
point may be coincident with one of the stakes, but must 
always be a hard point, so that the foot of the rod may stand 
at the same level for both readings. 




no. 55 



Field Work. To explain the method of obtaining the field 
notes necessary for making a profile, let 0, 1, 2, 3, • • •, 11 
(Fig. 55) represent a portion of a section to be leveled and 
plotted. Establish a bench mark at or near the beginning of 
the line, measure the horizontal length of the section, and set 
stakes 100 feet apart, numbering them 0, 1, 2, 3, and so on. 
Place the level at some point, as between 2 and 3, and take 
the reading of the rod on the bench = 4.832. Let PP' repre- 
sent the datum plane, say 15 feet below the bench mark ; then 

15 + 4.832 = 19.832 

is the height of the line of sight AB, called the height of the 
instrument, above the datum plane. 



290 



SURVEYING 



Now take the reading at = 5.2 = A, and subtract the 
same from 19.832, which leaves 14.6 = P, the height of the 
point above the datum plane. Next take sights at 1, 2, 3, 
3.40, and 4, equal, respectively, to 3.7, 3.0, 5.1, 4.8, and 8.3, and 
subtract the same from 19.832 ; the remainders 16.1, 16.8, 




Fig. 56 

14.7, 15.0, and 11.5 are respective heights of the points 1, 2, 
3, 3.40, and 4. 

Then, as it is necessary to change the position of the 
instrument, select a point in the neighborhood of 4 suitable 
as a turning point (t.p. in the figure), and take a careful 
reading on it = 8.480 ; subtract this from 19.832, and the 
remainder, 11.352, is the height of the turning point. 

Now carry the instrument forward to a new position, as 
between 5 and 6, shown in the figure, while the rodman 
remains at t.p.\ take a second reading on t.p. = 4.102, and add 
it to 11.352, the height of t.p. above PP'-, the sum 15.454 is 
the height of the instrument CD in its new position. 

Take sight upon 5, 6, and 7, equal, respectively, to 4.9, 2.8, 
and 0.904 ; subtract these sights from 15.454, and the results 
10.6, 12.7, and 14.550 are the heights of the points 5, 6, and 
7, respectively. 

The point 7, being suitable, is made a turning point, and 
the instrument is moved forward to a point between 9 and 10. 
The sight at 7 = 6.870, added to the height of 7 gives 21.420 
as the height of the instrument EF in its new position, The 



LEVELING 



291 



readings at 8, 9, 10, and 11, which are, respectively, 5.4, 3.6. 
5.8, and 9.0, subtracted from 21.420 give the heights of these 
points, namely, 16.0, 17.8, 15.6, and 12.4. 

Proceed in like manner until the entire section is leveled, 
establishing bench marks at intervals along the line to serve 
as reference points for future operations. The bench marks 
should be described with sufficient minuteness to enable any 
one not connected with the survey to locate them easily and 
unmistakably. A record of the work is given in the following 
tal^e: 



Station 


+ s. 


H.I. 


-S. 


H.S. 


Remarks 


B 


4.832 






15.0 


Bench on rock 20 ft. 







19.832 


5.2 


14.6 


south of 


1 






3.7 


16.1 




2 






3.0 


16.8 




3 






5.1 


14.7 


3 to 3.40 turnpike road 


3.40 






4.8 


15.0 




4 






8.3 


11.5 




t.p. 


4.102 




8.480 


11.352 




5 




15.454 


4.9 


10.6 




6 






2.8 


12.7 




7 


6.870 




0.904 


14.550 




8 




21.420 


5.4 


16.0 




9 






3.6 


17.8 




10 






5 8 


15.6 




11 






9.0 


12.4 




B 










Bench on oak stump 


12 










27 ft, N.E. of 12, 


etc. 




• 






etc. 



The first column contains the numbers or names of all the 
points on which sights are taken. The second column con- 
tains the sight taken on the first bench mark, and the sight 
on each turning point taken immediately after the instrument 



292 SURVEYING 

has been moved to a new position. These are called plus 
sights (-}- S.) because they are added to the heights of the 
points on which they are taken to obtain the height of the 
instrument given in the third column (H.I.). The fourth 
column contains all the readings except those recorded in the 
second column. These are called minus sights (— S.) because 
they are subtracted from the numbers in the third column to 
obtain all the numbers in the fifth column except the first, 
which is the assumed depth of the datum plane below the 
bench. The fifth column (H.S., height of surface) contains 
the required heights of all the points of the section named 
in the first column together with the heights of all benches 
and turning points. 

Making the Profile. Draw a line PP' (Fig. 56), to repre- 
sent the datum plane, and beginning at some point as P, lay 
off the distances 100, 200, 300, 340, 400 feet, and so on, to the 
right, using some convenient scale, say 200 feet to the inch. 
At these points of division erect perpendiculars equal in length 
to the height of the points 0, 1, 2, 3.40, 4, • • •, given in the 
fifth column of the above field notes, using in this case a larger 
scale, say 20 feet to the inch. Through the extremities of 
these perpendiculars draw the irregular line 0, 1, 2, 3, • • •, 11, 
and the result, with some explanatory figures, is the required 
plot or profile. 

The making of a profile is much simplified by the use of 
profile paper, which may be had by the yard or roll. 

If a horizontal plot is required, the bearings of the differ- 
ent portions of the section must be taken. Such a plot should 
be made, if it will assist in properly understanding the field 
work, or if it is desirable for future reference in connection 
with the field notes. Sometimes both the profile and the plot 
are drawn side by side on the same sheet; in this case, if 
the line leveled over is not straight, the profile will be longer 
than the plot. 



LEVELING 



293 



SECTION XX 



TOPOGRAPHIC LEVELING 



The principal object of topographic surveying is to show 
the contour of the ground. This operation, called topographic 
leveling, is performed by representing on paper the curved 
lines in which parallel horizontal planes at uniform distances 
from each other would meet the surface. It is evident that 
all points in the intersection of a horizontal plane with the 
surface of the ground are at the same level. Hence, it is 
necessary only to find points at the same level and join these 
to determine a line of intersection. 

The method commonly employed will be understood by 
reference to Fig. 57. The ground ABCD is divided into 
equal squares, and a numbered 
stake driven at each intersection. 
By means of a level and leveling 
rod the heights of the other sta- 
tions above m and D, the lowest 
stations, are determined. A plot 
of the ground with the intersect- 
ing lines is then drawn, and the P 
height of each station written as 
in the figure. 

Suppose that the horizontal 
planes are 2 feet apart ; if the 
first passes through m and Z>, the 

second will pass through p, which is 2 feet above m ; and 
since n is 3 feet above m, the second plane will cut the line 
mn in a point s determined by the proportion mn : ms = 3:2. 
In like manner, the points t, q, and r are determined. 

The irregular line tsp • • • qr represents the intersection of 
the second horizontal plane with the surface of the ground. 




294 SURVEYING 

In like manner, the intersections of the planes, respectively, 
4, 6, and 8 feet above m are traced. The more rapid the 
change in level the nearer these lines approach each other. 

SECTION XXI 
DRAINAGE SURVEYING 

Preliminaries. The locality to be drained should first be 
carefully reconnoitered, with the view of ascertaining the 
general feature of the land so as to enable the surveyor 
properly to locate the drains ; the beginning, route, and 
terminus of which should all be definitely planned. By the 
beginning of a drain is meant its highest point. 

Field Work. The field work is essentially the same for 
under drains and for open drains. The first thing is to 
establish the line of a drain. This includes the setting of 
stakes at intervals of from 50 feet to 100 feet, and also 
wherever there is an angle in the line; the bearings and 
lengths of the successive straight-line sections, beginning 
with the instrument set over the beginning of the drain ; 
and the designation by distances of the points of meeting 
of roads and land lines. Levels of the lines are then taken 
in accordance with the method described on pp. 289-291. If 
circumstances will permit, it is sometimes of advantage to 
have the leveling process go hand in hand with the estab- 
lishing of the line. 

Plot and Profile. If a considerable region is to be drained, 
a plot should be made of the entire tract, and on this plot 
should be drawn, in proper position, the lines of the drain 
and its branches. In a suitable place on the sheet should be 
noted the courses of the various sections of the drain and 
the number of linear feet belonging to each owner of land 
within the tract drained. A profile should also be made, 



LEVELING 



295 



as shown on page 292. From this profile inspection will 
determine whether a single grade will suffice, or whether a 
succession of different grades will be better. 



EXERCISE IX 

1. Find the difference of level of two places from the fol- 
lowing field notes: backsights, 5.2, 6.8, and 4.0; foresights, 
8.1, 9.5, and 7.9. 

2. Stake of the following notes stands at the lowest point 
of a pond to be drained into a creek ; stake 10 stands at the 
edge of the bank, and 10.25 at the bottom of the creek. Make 
a profile, draw the grade line through and 10.25, and fill out 
the columns H.G. and C, the former to show the height of 
grade line above the datum, and the latter, the depth of cut 
at the several stakes necessary to construct the drain. 



Station 


+ s. 


H.I. 


-s. 


H.S. 


H.G. 


C. 


Remarks 


B 


6.000 






25 






Bench on rock 









10.2 




20.8 


0.0 


30 ft. west of 


1 






5.3 






5.3 


stake 1 


2 






4.6 










3 






4.0 










4 






6.8 










5 


4.572 




7.090 










6 






3.9 










7 






2.0 










8 






4.9 










9 






4.3 










10 






4.5 










10.25 






11.8 











Horizontal scale, 2 ch. = 1 in. 
Vertical scale, 20 ft. = 1 in. 



296 



SURVEYING 



3. Find the difference in altitude between the highest 
point and the lowest point of the campus or of a field. 

4. Obtain the data necessary for a profile of a half mile of 
highway, and make the profile. 

5. Write the proper numbers in the third and fifth columns 
of the following table of field notes, and make a profile of the 
section. 



Station 


+ s. 


H.I. 


-s. 


H.S. 


Remarks 


B 


6.944 






20 


Bench on post 22 ft. 









7.4 




north of 


1 






5.6 






2 






3.9 






3 






4.6 






t.p. 


3.855 




5.513 






4 






4.9 






5 






3.5 






6 






1.2 







CHAPTER VI 
RAILROAD SURVEYING 

SECTION XX n 
LAYING OUT THE ROUTE 

Preliminary Survey. After it has been decided which of 
several feasible lines is the best, a preliminary survey for 
final location should be made. This should include, among 
other things, data referring to elevations, depressions, streams 
to be crossed, highways, buildings obstructing, character of 
soil, and natural resources affording materials for construc- 
tion j also data referring to proximity to towns, titles of 
land, rights of way, and so on. 

Establishing the Roadbed. When the general route of a 
railroad has been determined, a middle surface line is run 
with the transit. A profile of this line is determined, as on 
page 292. The leveling stations are commonly 1 chain 
(100 feet) apart. Places of different level are connected by 
a gradient line, which intersects the perpendiculars to the 
datum line at the leveling stations in points determined by 
simple proportion. Hence, the distance of each leveling 
station above or below the level or gradient line which 
represents the position of the roadbed is known. 

SECTION XXIII 

CROSS-SECTION WORK 

Excavations. If the roadbed lies below the surface, an 
excavation is made. Let ACBD (Fig. 58) represent a cross 

297 



298 



SURVEYING 



section of an excavation, / a point in the middle surface line, 
/' the corresponding point in the roadbed, and CD the width 
of the excavation at the bottom. The slopes at the sides are 
commonly made so that AA' = %A'C, and BB' = \DB\ When 
ff and CD are known, the points A, B, C, and D' are readily 
determined by a level and tape measure. 




/ 



D 



r 

Fig. 58 



D 



B 




If from the area of the trapezoid ABBA' the areas of 
the triangles AA'C and BB'D are deducted, the remainder 
is the area of the cross section. In like manner the cross 
section at the next station may be determined. These two 
cross sections, if similar, are the bases of a frustum of a 
quadrangular pyramid whose volume will be the amount of 
the excavation. In case the cross sections are not similar, 
the computations should be made by means of the Prismoidal 
Formula (Geometry, § 733). 

Embankments. If the roadbed lies above the surface, an 
embankment is made, the cross section of which is like that 
of the excavation, but inverted. 




Fig. 59 represents the cross section of an embankment which 
is lettered so as to show its relation to the excavation of Fig. 58. 



RAILROAD SURVEYING 



299 



SECTION XXIV 



CURVES 



Principles. When it is necessary to change the direction 
of a railroad it is done gradually by a curve, usually the 
arc of a circle. Let AF and 
AO (Fig. 60) represent two 
lines to be thus connected. 
Take any convenient length 
AlB = AE = t. The inter- 
section of the perpendicu- 
lars BC and EC determines 
the centre C, and the radius 
of curvature BC = r. The 
length of the radius de- 
pends on the angle A and the tangent AB. For, in the right 
triangle ABC, 

BC r 

tan BA C = —- > or tan \A = - • 

A Jl> t 




Fig. 60 



Hence, 



r = t tan % A . 



The degree of a railroad curve is the angle subtended at the 
centre of the curve by a chord of 100 feet. If D is the degree 
of a curve and r its radius, 

50 
sin % D = — j and r = 50 esc -J- D. 



For example, a 6° curve has a radius of 955.37 feet. 

Sometimes the topography of the route is such as to neces- 
sitate a successive series of curves of different radii, in which 
case the whole series of curves is called a compound curve, the 
principles involved being the same for each component as for 
a simple curve. 



300 



SURVEYING 



Methods of laying out the Curve. 1. Let Bm (Fig. 61) 
represent a portion of the tangent. It 
is required to find mP, the perpendicu- 
lar to the tangent meeting the curve 
at P. 

mP = Bn = CB — Cn. 
CB = r, 

Cn = ^Cp — Pn 




Fig. 61 




Hence, 



mP 



= Vr 2 - t\ 
= r 



Vr 2 - t\ 



2. It is required to find mP (Fig. 62) 
in the direction of the centre. 
mP = mC — PC. 
But mC = ^BC 2 + Bm 2 = Vr 2 + t\ 
Hence, 



Fig. 62 



mP = Vr 2 + t 2 — r. 



3. Place transits at B and E (Fig. 63). Direct the tele- 
scope of the former to E, 

and of the latter to A. Turn 
each toward the curve the 
same number of degrees, 
and mark P, the point of 
intersection of the lines of 
sight. P is a point in the 
circle to which AB and AE 
are tangents at B and E, 
respectively. 

4. If the degree D of the curve is given and the tangent 
BA at B (Fig. 64), place the transit at B and direct the tele- 
scope toward A. Turn off successively the angles ABP, PBP', 




Fig. 63 



RAILROAD SURVEYING 



301 



P'BP", •••, each equal to 
iD, and take BP, PP', P'P", 
■ • -, each 100 feet, the length 
of the tape. Then, P, P', 
P", • • • lie on the required 
curve. 

If the angle A and the 
tangent distance BA = t 
are given, D can be found 
from the formulas 

50 




Fig. 64 



sin £ D = — > and r = t tan \A. 
50 



Whence, sin \ D = — cot £ A. 



EXERCISE X 

1. The cross-section areas at five stations, 100 feet apart, 
of a railroad cut are, respectively, 576.8 square feet, 695.1 
square feet, 809.5 square feet, 652.0 square feet, and 511.7 
square feet. Compute the volume of material in this portion 
of the cut: (i) on the hypothesis that the cross sections are 
similar; (ii) on the hypothesis that they are dissimilar, the 
alternate cross sections being regarded as mid-sections. 

2. Find the radius of a curve of 1°, of 2°, of 3°, of 4°, of 5°. 

3. Two adjacent straight sections of a railroad form an 
angle of 148° 16'. They are joined by a curve touching each 
of them at the distance of 388 feet from the vertical point. 
Find the radius and the degree of the curve. 

4. Lay out a curve by the first or second method, and 
check the work by means of one of the transit methods. 



CHAPTER VII 

CITY SURVEYING 

SECTION XXV 

FIELD WORK 

Instruments. Since the principles in city surveying are 
essentially the same as those in land surveying, instruments 
of the same general character as the instruments already 
described may be used, except that in this class of work the 
ordinary compass and the chain are set aside. For the smaller 
cities, an instrument such as the surveyor's transit is suffi- 
cient in accuracy for the purposes of angle measurement and 
for leveling. However, when extreme accuracy is demanded, 
as in the case of large cities, specially made instruments 
should be used : a transit reading to 30 seconds, or even to 
10 seconds ; a high-grade Y level of at least 20-inch length ; 
and a standard tape, tested for sag and temperature. 

Streets. In most cases the city engineer must take the 
streets as he finds them. When a city has outgrown its 
original plan, if indeed it had any, sheer necessity may 
demand the location of additional streets or changes in 
existing streets. If a proposed town or city is to be laid 
out, the general contour of the ground and location of the 
site determine to a great extent the system of streets to be 
adopted. Experience has shown that wherever possible a 
rectangular system of street lines, with a few well-located 
diagonal streets, is the most satisfactory. Streets ordinarily 
vary in width from 50 to 100 feet, and each sidewalk from 7 

302 



CITY SURVEYING 



303 



Vincent 



St. 



r, 



Simpson 



o70' 

Ath let ic 
Field 



127.'5 




| 





























C-O-l-1- e-g-e Sewer St. jg- 

o' 



Campus 



Wal 



Pipp i " , o 



nut 



fcta-trer-tzr^™ r ^ c ^^S^t 



HI I 




S t 



Fig. 65 



to 15 feet. The principal improvements of streets are grad- 
ing, paving, setting curbs, laying sidewalks, constructing 
sewers, and laying water pipes. 



304 SURVEYING 

The field work necessary for all these may be included 
under the heads of leveling, locating lines, and locating 
points, which have already been described. 

Blocks and Lots. There is no established rule for the 
size of either blocks or lots. Fig. 65 gives some idea of their 
dimensions. The location of a block is described by refer- 
ence to the streets which bound it. A lot is described by 
number and block, or by number alone, or by giving the 
location and length of its bounding lines. The co-ordinate 
system of location of points, described on page 261, has 
much in its favor for use in city surveying. Monuments 
at points of reference and at intersections of streets and 
corners of lots should be of permanent character, and set 
with extreme care. 

SECTION XXVI 
0FFIC2 WORK 

Plots. Among the more important plots that should be 
prepared by the city engineer are a complete city map, drawn 
to scale, showing the streets and alleys, blocks and lots, with 
dimensions, and the location of railroads, street-car lines, sew- 
age system, water-pipe lines, and soon; a topographical map 
of the entire city, including as may be found desirable por- 
tions of the surrounding region ; a profile map of the streets. 
These are made from the field notes, which should be amply 
and carefully prepared. 

Records. No work of importance, whether done in the field 
or in the office, should fail to be recorded in some perma- 
nent form. Field notes, computations, plots, and copies of 
work specially prepared should be properly indexed and filed 
away. 



NAVIGATION 

CHAPTER I 
INTRODUCTION 

SECTION I 
DEFINITIONS 

Navigation is the art of conducting a ship from one port to 
another port. 

The True Course of a ship is the angle which her path 
makes with the true meridian. 

The Correct Magnetic Course is the angle which the ship's 
path makes with the magnetic meridian. 

When a ship is steered on one course, her path crosses the 
meridians at the same angle. Such a line is called the ship's 
track, rhumb line, or loxodromic curve. 

The Distance is the length of the path of the ship expressed 
in geographical or nautical miles. 

The two most important problems of navigation are : 

(1) To find the present position of the ship. 

(2) To determine the future course. 

The position (latitude and longitude) of a ship at sea may 
be found by two distinct methods : 

(1) By Dead Reckoning, which consists in keeping a record 
of the courses and distances sailed from a known point, and 
computing from these data the position reached. 

(2) By Nautical Astronomy, which consists in finding posi- 
tion by observations of the heavenly bodies, generally the sun. 

305 



306 NAVIGATION 

Problems of the first method except Great Circle Sailing are 
solved by Plane Trigonometry; problems of the second method 
are solved by Spherical Trigonometry. 




Fig. 1. The Mariner's Compass 



To find the position of a ship by observations of celestial 
bodies requires a knowledge of Nautical Astronomy and 
Spherical Trigonometry. When a ship is out of sight of 
land she is steered by the Mariner's Compass. 



DEFINITIONS 



307 



A Table of the Angles 
Which every Point and Quarter Point of the Compass makes with the Meridian 



North 


Points 

04 


o / // 


Points 


South 


N. by E. 


N. by W. 


2 48 45 
5 37 30 
8 26 15 
11 15 


0-| 

04 


S. by E. 


S. by W. 


N.N.E. 


N.N.W. 


14 
14 

1-3 

2 


14 3 45 
16 52 30 
19 41 15 
22 30 


14 
14 


S.S.E. 


S.S.W. 


N.E. by N. 


N.W. by N. 


24 

24 

24 

3 


25 18 45 
28 7 30 
30 56 15 
33 45 


24 
24 
24 

3 


S.E. by S. S.W. by S. 


N.E. 


N.W. 




36 33 45 
39 22 30 
42 11 15 
45 


34 
34 

r* 

4-1 

4-4 
4-3 

h 


S.E. S.W. 


N.E. by E. 


N.W. by W. 


M 
44 

4-1 
5 


47 48 45 
50 37 30 
53 26 15 
56 15 


S.E. by E. 


S.W. by W. 


E.N.E. 


W.N.W. 


11 


59 3 45 
61 52 30 
64 41 15 
67 30 


E.S.E. 


W.S.W. 


E. by N. 


W. by N. 


H 

64 


70 18 45 
73 7 30 
75 56 15 
78 45 


&4 

64 

7 


E. by S. 


W. by S. 


East 


West 


■4 
74 

1 ~i 

8 


81 33 45 
84 22 30 
87 11 15 
90 


74 
74 

7-3 
8 4 


East West 



The Card of the compass consists of a circular plate of mica 
covered with paper, the "circumference of which is usually 
divided into degrees. An inner circle is divided into thirty- 
two equal parts by radial lines called rhumb lines. The 
extremities of the rhumb lines are called points, or rhumbs. 

The principle upon which the points are named will be 
understood by inspecting a single quadrant. The point 
(N.E.) midway between NT. and E. is named by writing these 
letters together. The point (N.N.E.) midway between N. 
and N.E. is named by joining these letters. The first point 
to the east of north is named N. by E., and the first point to 



308 NAVIGATION 

the north of N.E. is named N.E. by N. In like manner, the 
points between E. and N.E. are named. 

Naming the points in order is called boxing the compass. 

One or more Magnetic Needles are attached to the lower sur- 
face of the card, parallel to the N. and S. line, having their 
N. and S. poles in the same direction as the corresponding 
points of the card. 

The card turns upon a pivot in the centre of a cylindrical 
metallic box called the Bowl, which is supported by gimbals 
in a square box. The bowl is hung within a brass hoop, and 
the hoop itself is hung at points 90° distant from the compass- 
bowl bearings, so that the card may retain a horizontal position 
in spite of the pitching and rolling of the ship. 

In some compasses the card almost floats in alcohol. This 
relieves the pivot of the greater part of the weight of the 
card and needles. 

The Lubber-Line is a distinct vertical line on the inner side 
of the bowl. 

It is often convenient to refer to the points of the compass 
as right or left of north or south. ^ Suppose an observer to 
stand at the centre of the card, facing north. Then, all points 
between N. and E. are right of north (R. of N.), and all points 
between N. and W. are left of north (L. of N.). In like man- 
ner, if the observer faces south, all points between S. and W. 
are right of south (R. of S.), and all points between S. and E. 
are left of south (L. of S.). 

Thus, W.S.W. is 6 points, or 67° 30', R. of S. 

N. by W. is 1 point, or 11° 15', L. of N. 
S.E. is 4 points, or 45°, L. of S. 

E. by N. is 7 points, or 78° 45', R. of N. 

The compass is placed in a box called the Binnacle. When 
properly placed, the lubber-line is next to the bow of the ves- 
sel, and a line passing through the lubber-line and the centre 
of the card is parallel to the fore-and-aft line of the vessel. 



VARIATION OF THE NEEDLE 309 

The Steering Compass is placed immediately before the 
helmsman. The Standard Compass is commonly placed near 
the middle of the vessel, in such a position as to be least 
affected by the iron used in the construction of the vessel. 
(See Sect. IV, p. 310.) 

SECTION II 

VARIATION OF THE NEEDLE FROM THE TRUE MERIDIAN 

This subject has been treated in Surveying. Variation is 
found for various places by astronomical observations. The 
navigator is provided with charts which give the variation 
at different places. 

SECTION III 

LOCAL ATTRACTION 

The disturbing effect upon the needle by iron outside of the 
ship (such as the iron in docks) is called local attraction. 

SECTION IV 
DEVIATION 

In a ship the needle will not always point to the magnetic 
north. 

Deviation includes the changes in the direction of the 
needle caused by the iron in the ship. 

Deviation is named easterly or westerly according as the 
north point of the needle is drawn to the eastward or to 
the westward of the magnetic north. 

The iron used in the construction of a ship consists of two 
kinds, soft iron and hard iron. Soft iron becomes instantly, 
magnetized to its full capacity when exposed to the influence 



310 NAVIGATION 

of any magnetized body, but loses its magnetism upon the 
removal of the magnet. This is called magnetism by induction. 

Hard iron does not become magnetized by ordinary induc- 
tion, but when magnetized retains its magnetism. Hence, 
this is called subjpermanent magnetism. A disturbance of the 
molecules of the iron by hammering or otherwise is required 
for it to receive or part with its magnetism. 

An iron ship may be correctly regarded as a large per- 
manent magnet, becoming so in the process of construction. 
The materials used in the construction do not by themselves 
retain magnetism permanently, but they acquire this property 
in a greater or less degree when united in the form of a ship 
and subjected to hammering, etc. 

Subpermanent magnetism depends upon the direction of the 
ship's head while building, and the direction of the dipping 
needle, that is, the variation of the needle, at the place of 
construction. It is found that if the ship was built head 
north, the deviation is westerly when the ship's head is east, 
and easterly when the ship's head is west. 

Induced magnetism is the magnetism induced in the soft 
iron in the vessel by the earth's magnetic force. 

The earth itself is a huge magnet, whose magnetic poles 
are entirely distinct from its poles of rotation. The north 
magnetic pole is N.N.W. of Hudson Bay; the south magnetic 
pole is to the south of Tasmania. The magnetic equator 
crosses the geographical equator near the meridian of Green- 
wich and near 168° W. longitude, making with the geographical 
equator an angle of about 12°. 

At all places on the magnetic equator a freely suspended 
needle takes a true horizontal position; that is, the dip of 
the needle is 0°. If the needle is taken to the north mag- 
netic pole, the red or north end of the needle will point 
vertically downward ; and at the south magnetic pole, verti- 
cally upward. 



DEVIATION 311 

Retentive magnetism is the temporary magnetism induced 
in an iron ship when her head is kept in one direction for 
some time. Retentive magnetism frequently remains for 
days after the cause is removed. 

Semicircular deviation is so named because it has the con- 
trary name and maximum value in opposite semicircles ; thus, 
if it is easterly on North, it is westerly on South. 

It is caused by subpermanent magnetism and the transient 
induced magnetism of vertical masses of soft iron in the ship. 
In wooden ships it disappears when the ship's correct mag- 
netic course is N. or S., and is greatest when the correct mag- 
netic course is E. or W. In iron ships the neutral points 
(points of no semicircular deviation) are in the direction of the 
ship's head and stern while building. The part due to sub- 
permanent magnetism remains the same in kind, either X. or S., 
but is different in amount in all latitudes, unless the ship is 
strained or subjected to mechanical violence. The part pro- 
duced by induced magnetism in vertical iron changes from 
X. to S. in different latitudes with the inclination of the 
dipping needle. It disappears at the magnetic equator, and 
is of contrary names on opposite sides of this equator. 

Quadrantal Deviation is caused by the transient induced 
magnetism of horizontal masses of soft iron in the ship. It 
disappears when the correct magnetic course is N., S., E., or 
W., and is greatest when the correct magnetic course is N.E., 
N.W., S.E., or S.W. ; hence the name quadrantal deviation. 
It has the same name in opposite quadrants and contrary 
names in adjacent quadrants ; thus, if it is easterly on N.E., 
it will be easterly on S.W., but westerly on S.E. and KW. 
It remains the same in all magnetic latitudes and is not 
appreciably affected by the lapse of time. 

The quadrantal deviation is compensated by soft-iron chains 
in boxes and horizontal cylinders of cast soft iron, or hollow 
cast-iron globes. 



312 NAVIGATION 

Retentive magnetism is corrected by over-compensating the 
quadrantal deviation when the course is E. and W., and by 
under-compensating when the course is N. and S. 

The semicircular deviation caused by subpermanent mag- 
netism is compensated by a system of magnets in the deck; 
that caused by induced magnetism by vertical soft-iron bars. 

The amount of deviation remaining after the compass has 
been adjusted may be found as follows : 

(1) By the known bearing of a distant object on shore. The 
ship is swung around so as to bring its head successively to 
the thirty-two points of the compass, and the bearing of the 
object on shore, as given by the standard compass when the 
ship's head is steadied on each point, is recorded in a table. 
The difference between the known correct magnetic bearing of 
the object and the successive bearings given by the standard 
compass gives the deviation for these points. A well-defined 
object should be selected so far distant that the diameter 
of the space through which the ship is swung will make no 
sensible difference in the real bearing from a central point. 
The Navy Department has established compass stations for 
convenience, with several points whose magnetic bearings are 
accurately determined. 

(2) By reciprocal bearings. If there is no suitable object 
on land, a second compass may be placed on shore, and the 
mutual bearings of this instrument and the standard compass 
on shipboard, when the ship's head is steadied on a certain 
point, taken at the same instant by preconcerted signal. The 
bearing given by the compass on shore is then reversed. 
The difference between this reversed bearing and the bearing 
given by the standard compass is the deviation for this point 
of the compass. In the absence of cargoes containing iron, 
wooden vessels going short voyages have their deviation 
nearly constant, in which case it is a good plan to write it 
on each point of the compass for ready reference. 



DEVIATION 313 

(3) By amplitudes and azimuths of a heavenly body. The 
deviation when at sea is determined from an azimuth. (See 
Sect. XXXI, p. 418.) 

The deviation is named as follows : If the correct magnetic 
bearing of the distant object (or the reversed bearing of the 
shore compass) is to the right of the reading of the standard 
compass, the deviation is easterly; if the correct magnetic 
bearing of the distant object (or the reversed bearing of the 
shore compass) is to the left of the reading of the standard 
compass, the deviation is westerly. 

A Deviation Table, or deviation card, is a tabulated statement 
of the amount of deviation for each point of the compass 
when the ship is on an even keel. 

When the ship heels under sail or is rolling heavily there 
is an additional error introduced for various degrees of heel- 
ing, called the heeling error, which must' be estimated and 
allowed for separately. Compensation is made by a vertical 
steel magnet placed exactly under the centre of the card 
when the vessel is upright. 

Example 1. A ship is headed X.E. by standard compass. 
A distant object on shore bears S. 70° TV. by the standard 
compass. The correct magnetic bearing of the object from 
the ship is known to be S. 89° TV. Find the deviation. 

Correct magnetic bearing = S. 89° W., or 89° R. of S. 

Observed bearing = S. 70° W., or 70° R. of S . 

Deviation (for the point N.E.) = 19° E. 

The deviation is east, because 89° E. of S. is to the right of 
70° E. of S. 

Example 2. A ship is headed E. by standard compass. 
The bearing of the compass on shore from the ship, as given 
by the standard compass, is S. 21° TV. The standard compass 
in the ship bears X. 8° TV. from the compass on shore. Find 
the deviation. 



314 



NAVIGATION 



Correct magnetic bearing (shore reversed) = S. 8° E., or 8° L. of S. 
Observed bearing = S. 21° W., or 21° R. of 8 . 

Deviation (for the point E.) = 29° W. 

The deviation is west, because 8° L. of S. is to the left of 
21° R. of S. 

SOLUTION BY DIAGRAM 

C.N. (Example 1) is the direction of the needle of the compass on ship. 

N.E. is the direction in which the ship's head is pointed. 

O is the direction of the object from the ship. 

Draw a line so that the line shall make with it an angle S. 89° W., 
and it will be the magnetic meridian. 

Since the needle of the compass on ship points to the right of magnetic 
north, the deviation is east, and is evidently the angle formed by the two 
lines M.N. and C.N. 



C.W. 




C.E. 



C.W. 




C.E. 



Fig. 2 

C.N. (Example 2) is the direction of the needle of the compass on ship. 

C.E. is the direction in which the ship's head is pointed. 

is the direction of the compass on shore from the ship. 

As the magnetic bearing of the compass on ship from the compass on 
shore is N. 8° W., the magnetic bearing of the compass on shore from 
the compass on ship is S. 8° E. ; hence, draw a line so that the line 
shall make with it an angle S. 8° E., and it will be the magnetic meridian. 

Since the needle of the compass on ship points to the left of magnetic 
north, the deviation is west, and is the angle formed by M.N. and C.N. 



LEEWAY 315 

SECTION V 
THE WIND 

The wind is given the name of the point of the compass 
from which it blows. 

Windward is the direction from which the wind blows. 

Leeward is the direction toward which the wind blows. 

When the wind is directly behind a ship it is said to be 
right aft, or right astern. The ship is then sixteen points from 
the wind. 

A ship is said to be on the starboard tack or the port tack 
according as the wind blows on the right side or the left side 
of the ship. 

When the wind blows at right angles to the side of a ship, 
the ship is said to have the wind on her beam. The ship is 
then eight points from the wind. 

When a ship is more than eight points from the wind, the 
wind is said to be abaft, the beam; when the ship is less than 
eight points from the wind, the wind is said to be before the 
beam. 

A ship is said to be close-hauled when she is as near the 
wind as she will lie ; that is, when she is headed as near as 
possible toward the point from which the wind blows. Few 
ships can lie closer than five points from the wind. 

SECTION VI 
LEEWAY 

The angle which the fore-and-aft line of a vessel makes with 
her track is called leeway. 

Let the arrow (Fig. 3) represent the direction of the wind, 
and CB the direction of the ship. Let CB represent the 




316 NAVIGATION 

distance the ship would go in a given time if she were not 
allowed to move sideways, and CD the distance she would 

go in the same time if she 
were not allowed headway. By 
a principle of mechanics, the 
ship would in the given time 
describe CA, the diagonal of the 
parallelogram of which CB and 
CD are adjacent sides. 

The angle BCA is the leeway. 
The amount of leeway de- 
pends upon the trim of the ship, her velocity through the 
water, the sails set, etc. It is sometimes found for a given 
ship, a given freight, and a given obliquity and velocity of 
the wind by special experiments. 

In making up the reckoning, leeway and heave of the sea 
must be allowed for according to judgment, as they vary in 
different vessels according to their build, and in the same 
vessel according to her draught and trim for the time being. 
It depends also upon the amount of wind and sea and the 
sails carried, so that no fixed rule for estimating it can be 
laid down. The leeway table usually seen in books on 
Navigation is, therefore, of but little value. 



SECTION VII 
REDUCING THE COURSES 

The Compass Course is the course steered. 

The true course is equal to the compass course corrected 
for variation, deviation, and leeway. 

From the preceding explanations, the following rules for 
changing from one course to the other are evident: 



REDUCING THE COURSES 317 

To find the true course from the compass course : 
Tr ... f Allow easterly variation to the right. 

V aViatlOTl < 

1 Allow westerly variation to the left. 

r. ... f Allow easterly deviation to the right. 
Deviation < 

I Allow westerly deviation to the left. 

T f Allow leeway to the left on the starboard tack. 

Leeway < J 

I Allow leeway to the right on the port tack. 

To find the compass course from the true course : 

Tr . ^. f Allow easterly variation to the left. 
Variation < J 

I Allow westerly variation to the right. 

r, ... f Allow easterly deviation to the left. 
Deviation -i 

\ Allow westerly deviation to the right. 

r f Allow leeway to the right on the starboard tack. 

Leeway < J ° 

L Allow leeway to the left on the port tack. 

Example. 1. The compass course is E.X.E.. wind S.E., 
leeway 3 points, variation 10° E., deviation 2° 30' E. Find 
the true course. 

Compass course (E.N.E.) is 6 points = 67° 30' R. of X. 

Leeway (starboard tack) is 3 points = 33° 45" L. 

Compass course (corrected for leeway) = 33° 45' R. of X. 

Variation and deviation (10° E. + 2° 30' E.) = 12° 30 r R. 

True course = N. 46° 15' E. 

Example 2. The compass course is W.S.W., wind S., lee- 
way 2^ points, variation 30° E., deviation 18° W. Find the 
true course. 

Compass course (W.S.W.) is 6 points = 

Leeway (port tack) is 2\ points = 

Compass course (corrected for leeway) = 

Variation and deviation (30° E. - 18° W. = 12° E.) = 
True course = 72° 23' L. of N. 

= N. 72° 23' W. 



67° 30' R. 


of S. 


28° V R. 




95° 37 r R. 


of S. 


84° 23 r L. 


of N. 


12° (T R. 





318 



NAVIGATION 



Example 3. The true course is N.E. by N., the vessel is 
on the port tack, leeway 10°, variation 1 point W., deviation 
5° 30' E. Find the compass course. 

True course (N.E. by N.) = 

Variation is 1 point W. = 

Correct magnetic course = 

Leeway (port tack) = 

True course (corrected for variation and leeway) = 
Deviation (for 35° R. of N. = N.E. by N. nearly) = 
Compass course = 29° 30' R. of N. 

= N. 29° 30' E. 



33° 45' R. of N. 
11° 15' R. 



45° 


0' 


R. 


of N. 


10° 


0' 


L. 




35° 


0' 


R. 


of N. 


5° 


30' 


L. 





SOLUTION BY DIAGRAM 

The solutions of Examples 1, 2, and 3 are clearly shown by referring 
to Fig. 4. 

T.N. is geographical north (True North). 

M.N. is direction of needle by variation (Magnetic North). 

C.N. is direction of needle by variation and deviation (Compass North). 

C. C. is direction in which ship's head is pointed (Compass Course). 

T.C. is direction in which ship goes owing to leeway (True Course). 



T.W. 




^fe; 



%> 



N*. 



T.E T.W- 



fo 



•T.E. 



Fig. 4 



Hence, in Example 1, the required course is evidently the angle formed 
by the two lines T.N. and T.C. 

In Example 2, the required course is evidently the angle formed by the 
two lines T.N. and T.C. 

In Example 3, the required course is evidently the angle formed by the 
two lines C. N. and C. C. 



THE LOG 



319 



EXERCISE I 



In the answers reckon 30" or over as V and reject less 
than 30". 



1 


Compass 
Course 


Wind 


Leeway 


Variation 


Deviation 


Trle Course 


S. 


E.S.E. 


Hpts. 


52° 0' W. 


2° O'E. 


Required 


2 


W.N.W. 


N. 


3 pts. 


42°0' E. 


18° 30' \V. 


Required 


3 


S.S.E. IE. 


S.W. iS. 


3* pts. 


2^ pts. E. 


1] pts. W. 


Required 


4 


Required 


S. by W. 


iPt- 


10°30 , E. 


19° O'W. 


S.79° W. 


5 


W. | N. 


N.N.W. 


If pts. 


8°30'E. 


15°35'E. 


Required 


6 


E. \ N. 


N.N.E. 


2i pts. 


13° O'W. 


20° O'E. 


Required 


7 


Required 


N. by W. 


ipt. 


14° O'E. 


19° O'E. 


S. 85° E. 


8 


W. 


N.N.W. 


lipts. 


18°30'E. 


21° O'W. 


Required 


9 


E. i S. 


N.N.E. IE. 


2fcpts. 


21° O'E. 


4° O'W. 


Required 


10 


Required 


N. by W. 


2f pts. 


2 pts. W. 


3£ pts. E. 


E. by S. \ S. 


11 


Required 


N. E. 


3 \ pts. 


2f pts. E. 


Upts. E. 


N. by W. 


12 


Required 


S.S.W. 


2i pts. 


2J pts. E. 


fpt. E. 


N.N.W. 


13 


Required 




pts. 


7°0 / W. 


15° O'W. 


S. (54° E. 


14 


Required 




pts. 


6°0'E. 


20° O'W. 


N. 44° W. 


15 


N. 65° W. 




pts. 
pts. 


10°0 / E. 


3°0'E. 


Required 
Required 


16 


S. 15° W. 




6°0'W. 


18°0'E. 


17 


S. 18° E. 




pts. 


25°0 / E. 


10°0'E. 


Required 


18 


N. 30° E. 


S. by W. 


H P ts - 


12°0 , E. 


10° O'W. 


Required 



SECTION VIII 
THE LOG 

The rate at which a vessel moves 
through the water is commonly 
determined by the log. 

The Common Log consists of the 
log-chip (often called simply the 
log), the log line, and the sandglass. 

The Log-Chip (Fig. 5) is a flat piece of wood in the form of 




Fig. 5 



320 



NAVIGATION 



a sector, of about 5 inches radius, having its arc loaded with 
lead to make it float in an upright position, so that the vertex 
shall be just above the water. 

The Log Line, to which the log-chip is attached by three 
strings, is an ordinary cord about 150 fathoms long. The 
stray line is the part of the log line next the log-chip, and is 
commonly about two thirds the length of the ship. The end 
of the stray line is marked by a rag tied into the line. The 
remainder of the line is divided into equal parts called knots, 
and each knot is divided into tenths. Beginning at the end 
of the stray line, the end of the first knot is 
marked by a piece of cord having one knot 
tied in it ; the end of the second knot is marked 
by a cord with two knots tied in it ; the eod 
of the third knot is marked by a cord with 
three knots tied in it; and so on. 

The stray line allows the log to settle 
in the water clear of the ship before the 
measurement of the rate begins. 

A Sandglass (Fig. 6) of 30 seconds, 28 
seconds, or 14 seconds may be used. 

The knot is made the same part of a 
nautical mile that the time required for 
the sand to run through the glass is of an 
hour. 




Fig. 6 



To find the length of a nautical mile. 

If the radius of the earth is 3960 statute miles, the circum- 
ference of the earth, 2 nrr, is 2 x 3.1416 x 3960 statute miles 
= 24881.4720 statute miles. 

Length of 1° = 24881.4720 statute miles -- 360 = 69.1152 
statute miles. - fiQ11?i9 

Length of 1' = 1 nautical mile = — '— — X 5280 feet = 
6082 feet. 



THE LOG 321 

The knot corresponding to a 30-second glass would be oOf 
feet, determined by the following proportion, using 6080 feet 
as a nautical mile : 

3600 seconds : 30 seconds = 6080 feet : 50f feet. 

As a vessel generally overruns her reckoning, the knot 
corresponding to a 30-second glass is made 50 feet. The 
proportion would give 47^f feet as the length of the knot 
corresponding to a 28-second glass, but in practice it is made 
46 feet 8 inches. 

The log line is wound upon a reel (Fig. 7). 

The operation of heaving the log is as follows : The officer 
throws the log-chip 
into the water on 
the lee side of the 
vessel. It remains 
nearly stationary in 
the water while the 
line is reeled off as 
the vessel moves for- 
ward. The instant 
that the mark indi- 
cating the end of the stray line passes over the side of the ship 
the glass is turned. The operator then counts the number of 
knots run out during the time required for the sand to run 
through the glass. The line is given a quick jerk, which 
detaches the two lower of the three cords by which the log- 
chip is attached to the line, the log falls flat, and is readily 
drawn in. 

Since each knot is (approximately) the same part of a 
geographical mile that the time is of one hour, the number of 
knots run out while the sand is running through the glass 
indicates the rate of the vessel in geographical miles per 
hour. 




Fig. 7 



322 



NAVIGATION 



The Patent Harpoon Log, one form of which is shown in 
Fig. 8, consists of a brass cylinder, which is towed through 
the water after the vessel by a line about 50 fathoms long. 
The cylinder consists of two parts. The hinder part, or 




Fig. 8 



rotator, is made to revolve by four curved blades as the log 
is drawn through the water ; the forward part is kept from 
revolving by two straight blades, and contains a registering 
apparatus connected with the rotator, which shows the distance. 




Fig. 9 



The Patent Taffrail Log, one form of which is shown in 
Fig. 9, consists of a registering apparatus which is perma- 
nently attached to the taffrail of the vessel, and a rotator 



THE L(Xi 323 

which is towed through the water by a cord connected with 
the register. The revolutions of the rotator are communicated 
to the register by the cord. 

The Ground Log consists of the ordinary log line with a hand 
lead substituted for the log-chip. When it is hove, the lead 
sinks to the bottom and remains stationary. The line reeled 
off gives the speed over the bottom. This log is not affected 
by tides or currents. It may be used in shallow water. 

The rate of paddle steamers can be estimated from the num- 
ber of revolutions of the paddle wheels quite as accurately as 
by the best patent log, especially in smooth water. 

The Different Logs compared 

The best results are obtained by the use of the harpoon log, 
if it is in good order; but the rapid motion of parts of the 
registering apparatus soon puts the best patent log out of 
order. Hence, fast ocean steamers ordinarily use the common 
log (hove every two hours) when in the open sea, and both the 
common and the patent log when nearing land. 

The taffrail log is the most convenient, since the register- 
ing apparatus can be examined at any time without taking 
the rotator from the water ; and if the rotator is lost, it may 
be replaced at small cost, whereas if the harpoon log is lost 
(a not infrequent occurrence), the cost of replacing it is con- 
siderable. Although the taffrail log is growing in favor, it is 
considered not as trustworthy as the harpoon log. 

Example 1. What should be the length of the knot on 
the log line when the sandglass runs out in 27 seconds, the 
nautical mile being 6080 feet ? Ans. 45| feet. 

Example 2. A log line is knotted every 59£ feet. Find 
what glass must be employed, the nautical mile being 6080 
feet. Ans. 35 seconds. 



324 



NAVIGATION 



SECTION IX 



CURRENTS 



The set of a current is its direction, or the point of the 
compass toward which it is moving. 

The drift of a current is its rate in miles per hour. 

Example 1. If a log is thrown from a ship in a current, 
will it show the rate of sailing or the rate over the bottom ? 

Ans. In a current, the log gives the speed through the 
water and not the speed over the bottom, since the log as well 
as the ship is carried by the current. 

The set and drift of a current may be ascertained by taking, 
at a short distance from the ship, a boat which is kept from 
being carried by the current by letting down to a consider- 
able depth a heavy weight attached by a rope to the stern of 
the boat. The method of allowing for the effect of a current 
upon a ship will be explained hereafter. 

Example 2. The wind is carrying a ship at the rate of 
15 miles an hour ; a current gives her a lateral velocity of 
10 miles an hour. Find the velocity of the ship over the 



/ ^^ 


B 




R 




/>"Y! 






yS 1 ^\ 




AN 
Fig. 10 





N' 
Fig. 11 


A 



bottom when the direction in which the current carries the 
ship makes (1) an angle of 60° and (2) an angle of 150° 
with the direction in which the wind is carrying the ship. 
(See Sect. VI, p. 316.) - 

Let OA (Figs. 10 and 11) represent the distance and direc- 
tion the ship would be carried by the wind in one hour if 



CURRENTS 



325 



there was no current, and let OB represent the distance and 
direction the ship would be carried by the current in one 
hour if there was no wind. Then OR represents the distance 
and direction the ship is carried in one hour by the wind and 
current acting together. 

~OR 2 = {OA+AN)* + NR 2 . OR 2 = (OA-NA)* + NR 2 . 

AN=ARcosW° = 10 x i = 6. If A = ^Lficos30°=10 X *V3 = 5 V3 

NR =^ J Rsin60 o =10xiV3=5\ / 8. NR = ARsin 30° = 10 x \= 5. 



OR 2 = (15 + 5) 2 + (5 V8)2 = 475. 
OB = V475 = 21.8. 

Therefore, the velocity of the ship 
over the bottom is 21.8 miles an 
hour. 



OR 2 = (15- 5 Vg)2 + 52 

= 325 - 150 V3. 
OR = V325 - 1* 

Therefore, the velocity of the ship 
over the bottom is 8.07 miles an 
hour. 



CHAPTER II 
THE SAILINGS 

SECTION X 
DEFINITIONS 

The Log Book is a record of the ship's course, containing 
the data from which the navigator establishes the ship's 
position by dead reckoning. 

The Difference of Latitude of two places is the arc of a 
meridian included between the parallels of the two places. 

The Middle Latitude of two places is the latitude of a 
parallel midway between the two places. 

The Difference of Longitude of two places is the arc of the 
equator included between the meridians of the two places. 

The Departure is the distance measured on a parallel between 
the meridian left and the meridian reached. 

SECTION XI 

NOTATION 

C = the true course. 
D = the distance in geographical miles. 
L = latitude in general. 
V = the latitude left (latitude from). 
L" = the latitude arrived at (latitude in). 
L m = the middle latitude. 
L d = the difference of latitude. 
326 



PLANE SAILING 



327 



A. = longitude in general. 

V = the longitude left (longitude from). 

V = the longitude arrived at (longitude in). 
\ d = the difference of longitude. 

p = the departure. 



SECTION XII 

PLANE SAILING 

Given the course and the distance, to find the difference of 
latitude and the departure : 



Case I 

When the distance is so small that the curvature of the earth 
may be neglected. 

Let CA (Fig. 12) be the distance sailed, CB the meridian 
through C, and AB the parallel through A. 
Then, angle ACB = C, 

CA = D, 

CB = L d , 

AB=p. 
By Plane Trigonometry, 



sin C = ^ , 






D 

D sin C, L 
V 



cos C = ^ ' 



tan C = 2- 



sin C 



D 



- D cos C, p = L d tan C ; 

-^-, L = P 

cos C d tan C 




Fig. 12 



Case II 

When the distance is so great that the curvature of the 
earth cannot be neglected. 



328 



NAVIGATION 



Let PCD (Fig. 13) represent a portion of the earth's sur- 
face, P the pole, CA the ship's track 
between C and A, PC and PD the 
meridians through C and A, CD and 
BA the parallels through C and A. 

Let CJ. be divided into parts Ch, 
hj, •••, so small that the curvature 
of the earth may be neglected for 
these distances ; and let gh, ij, • • • , 
be intercepts on the parallels corre- 
sponding to Ch, hj, • • • 

PjA = ---=C. 




Fig. 13 



Angle PC A = PhA 

By Plane Trigonometry, 

gh = Ch sin C 
ij = hj sin C, 



Hence, 
Again, 



gh + ij H = (Ch + /y H ) sin C. 

p = D sin C. 

Cg = CA cos C, 
hi = hj cos C, 



(a) 



.*. Cg + hi-\ = (Ch + hj -\ )cos C. 

Hence, L d = D cos C. 

From (a) and (b), 



0>) 



sin C = - , 
D 




cosC = — , 
D 




tanC = ^-, 


p = D sin C, 


f[l] 


L d = D cos C, 


[ffl 


p = L d tan C, 


D= P . 

sinC J 




D= L *. 

cosC 




d ~ tanC 



These formulas are the same as in Case I ; but in Case II 
p is the sum of the partial departures, each being taken in the 



PLANE SAILING 329 

latitude of its triangle, and is evidently greater than BA and 
less than CD. 

Plane Sailing, as here treated, may be denned as the method 
of establishing the relations between course, distance, differ- 
ence of latitude, and departure, on the supposition that small 
portions of the earth's surface may be regarded as plane. 

The several problems are readily solved by formulas [1], 
[2], and [3], but the method by inspection is commonly used 
by navigators. A number of examples will be solved by this 
method in order to illustrate the use of the Traverse Tables. 

Example 1. A ship sails from latitude 40° 20' N. on a 
N.N.E. course for 92 miles. Find the departure and the 
latitude in. 



Solution I. By computation. The true course N.N.E. = N. 22° 30' E 


To find the departure : 


To find the difference of latitude 


By [1], p = D sin C. 


By [2], L d = D cos C. 


D = 92 


logD= 1.96379 


logD =1.96379 


C = 22° 30' 


log sin C = 9.58284 


log cos C = 9.96562 


logp = 1.54663 


logXj= 1.92941 


.-. p = 35.2 miles. 


.-. Ld = 85.0 miles 




= 1°25 / N. 




U = 40°20'N. 






.-. L" = 41° 45' N. 



Solution II. By inspection of the Traverse Table. Since the hypote- 
nuse and an acute angle of a right triangle are given to find the sides, 
the Traverse Table may be used. Corresponding to the course N.N.E. 
(2 points), and distance 92 miles, Table VIII* gives latitude = 85 and 
departure = 35.2. The difference of latitude, 85 miles, is 1° 25'. 

Example 2. A ship sails S.W. by W. 488 miles. Find the 
difference of latitude and the departure. 

By inspection of the Traverse Table. S.W. by W. = 5 points. In 
Table VIII, at the bottom of page 87, we find 5 points; hence, we look 
at the bottom of the page for the designations of the columns. The table 

* Wentworth and Hill's Logarithmic and Trigonometric Tables. 



330 NAVIGATION 

extends only to a distance of 300 ; hence, we find the latitudes and depar- 
tures corresponding to the distances 300 and 188, and take their sum. 

For distance 300 latitude = 166.7 departure = 249.4 

« 188 " = 104.4 » = 156.3 

" 488 " =271.1 " =405.7 

Or, we may divide the given distance by any convenient number, and 
find the latitude and departure corresponding to the quotient (the course 
remaining unchanged), then multiply each of these by the divisor of the 
distance. 

488 -*- 2 = 244. 

For distance 244 latitude = 135.6 departure = 202.9 

-1 —1 _? 

For distance 488 latitude = 271.2 departure = 405.8 

The results differ slightly by the two methods ; but this 
small discrepancy is of no importance in practice. (See 
Sect. XVIII, p. 366.) 

Example 3. Course 6^ points, distance 21.7 ; required the 
difference of latitude and the departure. 

By inspection of the Traverse Table. 6£ points is found at the bottom 
of p. 81, Table VIII. Corresponding to distance 217 (21.7 x 10) the differ- 
ence of latitude is 63.0, and the departure 207.7. Dividing each of these 
numbers by 10, the required difference of latitude is found to be 6.3, and 
the departure 20.8. 

Example 4. If a ship runs S.E. 'by S. from 1° 44' north 
latitude, and is then by observation in 2° 46' south latitude, 
what are the distance and the departure ? 

S.E. by S. = 3 points. 1° 44' N. 
2° 46 7 S . 
4° 30' = 270 miles = difference of latitude. 

On p. 87, Table VIII, we find 3 points at the top ; but the greatest 
difference of latitude on this page is 249 ; hence, we divide the given 
difference of latitude by 2. 270 -4- 2 = 135. The nearest difference of 
latitude to 135 is 134.7, opposite which the distance is 162 and the 
departure 90. Multiplying each of these numbers by 2, the required 
distance is found to be 324 and the departure 180. 



PLAXE SAILING 



331 



Example 5. A ship sails N. 25° W. until the departure is 
98 miles. Find the distance and the difference of latitude. 



Solution I. By computation. 
To find the distance : 


To find the difference of la 


By[r 

C = 25° 
p = 98 


|, D = j) esc C. 

\ogp = 1.99123 
log esc C = 0.37405 
log D = 2.36528 
.-. D = 231.9. 


By [3], L d =p cote. 

logp = 1.99123 
log cot C = 0.33133 
log L d = 2 32256 
..L d = 210.2. 



Solution II. By inspection of the Traverse Table. Since the course 
is given in degrees, we turn to Table IX. At the top of p. 116 we find 
25°, and opposite departure 98 we find the required distance 232, and the 
difference of latitude 210.3. 

Example 6. Find the course and the distance correspond- 
ing to a difference of latitude of 696 miles and a departure of 
186 miles, the course lying between N. and E. 

Solution I. By computation. 

To find the course : To find the distance : 



By [3], tan C 



1L 
L d 

p =186 I logp = 2.26951 

L d = 696 I colog L d = 7.15739-10 
log tan C - 9.42690 

.-. C = N. 14°58'E. 



By[l], d = 



sin C 



logp = 2.26951 
colog sin C = 0.58795 
log 2) = 2.85746 
.-. D = 720.2. 



Solution II. By inspection of the Traverse Table. Since the given 
difference of latitude exceeds the limits of the tables, we divide each of 
the given numbers by 3 : 696 -=- 3 = 232 ; 186 -5- 3 = 62. If the course is 
desired in degrees, we now turn to Table IX and look for a difference 
of latitude of 232 opposite a departure of 62. On p. 106 the nearest 
approximation is found. At the top of this page the angle is 15° ; hence, 
the required course is N. 15° E. If the course is desired in points, we 
turn to Table VIII. The nearest approximation is found on p. 80, at 
the top of which stands 1^ points. Hence, the required course is N. by 
E. \ E. The distance is 720 (= 240 x 3) miles. 

Note. The answers given at the end of this volume are obtained by 
computation. The student should also solve by inspection. 



332 



NAVIGATION 



EXERCISE II 





L' 


L" 


C 


I) 


P 


1 


49° 57' N. 


Required 


S.W. by W. 


488.0 


Required 


2 


1° 45' N. 


Required 


S.E. by E. 


487.8 


Required 


3 


3° 15' S. 


Required 


N.E. by E. f E. 


449.1 


Required 


4 


2° 10' S. 


Required 


N. by E. 


267.0 


Required 


5 


41° 30' N. 


Required 


s.s.w. 


295.5 


Required 


6 


21° 59' S. 


24° 49' S. 


Required 


360.0 


Required 


7 


2° 9' S. 


3° 11' N. 


Required 


354.0 


Required 


8 


1° 30' N. 


0° 26' S. 


S. by W. 


Required 


Required 


9 


40° 17' N. 


37° 6'N. 


S. by W. \ W. 


Required 


Required 


10 


38° O'N. 


Required 


S.W. by W. 


Required 


48.2 


11 


18°25 / N. 


Required 


S.W. by W.f W. 


Required 


65.1 


12 


50° 18' N. 


54° 48' N. 


Required 


299.0 


Required 


13 


32° 30' N. 


19° 59' N. 


Required 


812.0 


Required 


14 


2° 8'S. 


Required 


N. 11° E. 


500.0 


Required 


15 


20° 21' S. 


Required 


N. 20° E. 


402.0 


Required 


16 


40° 25' S. 


Required 


N. 87° E. 


240.0 


Required 


17 


20° 48' N. 


17° 13' N. 


Required 


Required 


289.2 W. 


18 


.51° 45' N. 


53° 11' N. 


Required 


Required 


128.0 E. 


19 


0° 20'' S. 


0° 18' N. 


Required 


Required 


142.7 E. 


20 


40° 20' N. 


41° 37' N. 


Required 


Required 


52.6 W. 

1 



SECTION XIII 



PARALLEL SAILING 



Parallel Sailing is that case of sailing in which the track is 
on a parallel of latitude. 

Let P (Fig 15) represent the pole, O the centre of the earth, 
A and B two places on the same parallel AB, PAE and PBQ 
meridians through A and B intercepting EQ on the equator. 
Let AD and BD be perpendicular to PO. Draw AO. AB is 
the departure, or meridian distance Begard the earth as 
spherical. 



PARALLEL SAILING 



333 



AE = BQ = EGA = GAD = latitude of A and B. 
In the right triangle AGD, 

AJ) 
R' 



AD AD 

cos ODA = — - = —z- ? or cos L 



AG R 
.'. AD = R cos L. 
By Geometry, EQ : AB = EO : A D, 
or X d : p "== R : R cos Z 

.*. p = \ d cosL,") 
and \ d = p sec L J 



m 




Fig. 14 




These relations may be graphically represented by a right 
plane triangle. 

Now p = \ d cos L. Therefore, \ d is the hypotenuse of the 
right triangle of which L is an acute angle and p the leg 
adjacent, as in Fig. 16. 

Since p = X d cos L, the meridian distance 
varies as the cosine of the latitude. 

Let \ d = l° = 60' = 60 miles; then the 
above formula becomes p = 60 cos L. 

Hence, to find the number of geographical 
miles in 1° of longitude at any place, mul- 
tiply 60 by the cosine of the latitude. 




334 



NAVIGATION 



The number of statute miles in a degree of longitude may- 
be found by multiplying the numbers in the table by 
or by using the formula p = 69.115 cos L. 



60 



Table showing the Number of Geographical Miles in a Degree 
oe Longitude for Each Degree of Latitude 



L 


Miles 


L 


Miles 


L 


Miles 


L 


Miles 


L 


Miles 


L 

76° 


Miles 


1° 


59.99 


16° 


57.67 


31° 


51.43 


46° 


41.68 


61° 


29.09 


14.51 


2° 


59.96 


17° 


57.38 


32° 


50.88 


47° 


40.92 


62° 


28.17 


77° 


13.50 


3° 


59.92 


18° 


57.06 


33° 


50.32 


48° 


40.15 


63° 


27.24 


78° 


12.48 


4° 


59.85 


19° 


56.73 


34° 


49.74 


49° 


39.36 


64° 


26.30 


79° 


11.45 


5° 


59.77 


20° 


56.38 


35° 


49.15 


50° 


38.57 


65° 


25.36 


80° 


10.42 


6° 


59.67 


21° 


56.01 


36° 


48.54 


51° 


37.76 


66° 


24.41 


81° 


9.38 


7° 


59.56 


22° 


55.63 


37° 


47.92 


52° 


36.94 


67° 


23.44 


82° 


8.35 


8° 


59.42 


23° 


55.23 


38° 


47.28 


53° 


36.11 


68° 


22.48 


83° 


7.32 


9° 


59.26 


24° 


54.81 


39° 


46.63 


54° 


35.27 


69° 


21.50 


84° 


6.28 


10° 


59.09 


25° 


54.38 


40° 


45.96 


55° 


34.41 


70° 


20.52 


85° 


5.23 


11° 


58.89 


26° 


53.93 


41° 


45.28 


56° 


33.55 


71° 


19.53 


86° 


4.18 


12° 


58.69 


27° 


53.46 


42° 


44.59 


57° 


32.68 


72° 


18.54 


87° 


3.14 


13° 


58.46 


28° 


52.97 


43° 43.88 


58° 


31.79 


73° 


17.54 


88° 


2.09 


14° 


58.22 


29° 


52.47 


44° 43.16 


59° 


30.90 


74° 


16.54 


89° 


1.05 


15° 


57.95 


30° 


51.96 


45° 42.43 

i 


60° 


30.00 


75° 


15.53 


90° 


0.00 

1 



Example 1. A ship in latitude 42° changes her longitude 
3° 20' by sailing on this parallel. Find her departure. 



Solution I 

By [4], p = \ d cos L. 
p = 60x3|x cos 42° 



Solution II 

The table gives for L 



42 c 



1° of \ = 44.59. 
3i° of X = 31 x 44.59 
= 148.6 miles. 



= 60 x 3| x 0.7431 

= 148.6 miles. 

Solution III. By referring to Fig. 16, it will be seen that the given 
latitude corresponds to course, and difference of longitude to distance. 
Hence, p is the latitude in the Traverse Table corresponding to course 
42° and distance 60 x 3£ = 200, which gives 148.6 miles, as before. 



PARALLEL SAILING 



335 



Example 2. A ship sails 176.2 miles due west from the 
Lizard, in latitude 49° 57' N. and longitude 5° 14' W. Find 
the longitude in. 

Solution I 

By [4], \ d =psecL. 

p = 176.2 \ogp = 2.24601 

L = 49° 57' log sec L = 0.19148 
X' =5° 14' W. log X,, = 2.43749 

\q = 4°34'W . .-. X^ = 273.8 miles 

X" = 9° 48' W. = 4° 34' VV. 



SOLUTIOK II 

From the table, 

l°Xfor49°Z = 39.36, 
PXfor 50° L = 38.57. 

By interpolation. 

1° X for 49? 57' L = 38.61. 
176.2 -38.61 = 4.56. 

X,, = 4.56° = 4° 34'. 



EXERCISE III 



L 


P 


A' 


A 


1 55° 55' 


Required 


2° 10' W. 


12°52'E. 


2 


52° 00' 


Required 


0° 59' W. 


2° 24' E. 


3 


61° 25' 


Required 


179° 20' W. 


176° 52' E. 


4 


56° 00' 


Required 


3° 12' W. 


4° 8' E. 


5 


80° 00' 


Required 


10° 0' W. 


17° 41' W. 


6 


60° 00' 


204.0 E. 


160° 2' E. 


Required 


7 


51° 28' 


70.9 E. 


32° 7' W. 


Required 


8 


64° 16' 


265.7 W. 


170° 0' W. 


Required 


9 


46° 37' 


352.0 E. 


163° 42' E. 


Required 


10 


39° 57' 


398.0 W. 


4° 8' W. 


Required 



11. From latitude 32° 3' S., longitude 179° 45' W., a ship 
makes 54 miles west (true). Eequired the longitude in. 

12. From latitude 35° 30' S., longitude 27° 28' W., a ship 
sails east (true) 301 miles. Eequired the longitude in and 
the compass course ; variation If points E., leeway 4; point, 
the ship being on the port tack, deviation 8° 50' E. 

Note. In changing from the true course to the compass course, apply 
variation and leeway, then deviation, which is supposed to be given for 
the course to which it is directly applied. 



336 



NAVIGATION 



SECTION XIV 



MIDDLE LATITUDE SAILING 

Middle Latitude Sailing is a combination of Plane and Par- 
allel Sailings, chiefly for the purpose of determining differ- 
ence of longitude. It is based on the supposition that the 
departure in Plane Sailing is equal to the distance between 
the meridians passing through the extreme points of the rhumb 
line, measured on the middle parallels between these points. 

Let CA (Fig. 18) represent the ship's track, CD and BA 
parallels, and PBC and PAD meridians through C and A, 
respectively. Let EF be the parallel midway between C and A . 





Fig. r, 



Fig. 18 



It is shown in Plane Sailing that p is less than CD and 
greater than BA. It is nearly equal to EF, and exactly equal 
to an intercept a little nearer to the pole. than EF, for it 
is plain that the middle latitude distance receives a much 
greater accession than the departure if the track CA cuts the 
successive meridians at a very small angle. 

Hence, L m = -J (£' -f L"), nearly, if both places are on the 
same side of the equator. 

In Middle Latitude Sailing p is first found as in Plane Sail- 
ing, then X cl is found, on the supposition that X d is equal to 



MIDDLE LATITUDE SAILING 



337 



the change of longitude which the ship would have made by 
sailing a distance p on a parallel in latitude L m . 

If we substitute for p its value in the formulas, 

In Parallel Sailing, 

% [ 4 ]> P = x d c o s L m> K = V sec L m . [5] 

In Plane Sailing, 

By [1] and [3], p = D sin C, tan C = %- • 



Hence, by combining, 



tanC = 



K cos L„ 



X d = D sin C sec L m . 



[6] 



The formula L m = ^ (L' + L") may be used without sensible 
error in low latitudes when the course is not less than about 
45° and both places are on the same side of the equator. 

When the places are on different sides of the equator and 
the distance not great, p = \ d may be used, since the meridians 
near the equator are sensibly parallel ; but when the distance 
is considerable the two portions of the track on different sides 
of the equator should be treated separately. 

Let CA (Fig. 19) represent the rhumb line crossing the 
equator EQ at B. Let CE and AQ 
represent meridians. 

For the rhumb CB the middle 
latitude is -J- CE; hence, EB (the 
difference of longitude of C and B) 
may be found. For the rhumb 7L1 
the middle latitude is % AQ; hence, 
BQ (the difference of longitude of 
B and A) may be found. 

Then, X d = EB + BQ. 




Fig. 19 



338 



NAVIGATION 



In high latitudes, or in any latitude, if the course is less 
than about 45° and the distance great, 

d being a small arc. By means of a small table of correc- 
E tions, the imperfections of the middle lati- 
tude method may be removed and the 
results rendered accurate. This correction 
of the middle latitude is seldom used in 
practice. 

The relations of the elements of Middle 
Latitude Sailing may be graphically repre- 
sented by combining the triangles of Plane 
and Parallel Sailings, as in Fig. 20. 

ACB=C, AB=p, 

CA = D, BAE = L m , 




CB = L, 



AE = A,. 



Fig. 20 

Example 1. A ship sails from Sandy Hook light, in lati- 
tude 40° 28' K, longitude 74° W., on an E.S.E. course, 62 



Find the latitude in and the longitude in. 



miles. 

Solution 1. By computation 
To find the latitude in : 



By [2], L d = D cos C. 



D = 62 


log£>= 1.79239 


C = 67° 30' 


log cos C = 9.58284 


\ogL d = 1.37523 


.-. 2^ = 23.7' = 24' S. 


£' = 40°28'N. 


L d = 24' S. 


L // = 40 o 4 / N 


To find the middle latitude : 


U =40°28'N. 


£"=40° 4'N. 


2) 80° 32' 


L in - 


= 40° 16' N. 



To find the longitude in : 

B y [ 7 ]> K = D sin C sec L m . 

logD= 1.79239 

log sin C= 9.96562 

log sec L m = 0.11745 

log \ d = 1.87546 

.-. X d = 75.r 

= 1° 15' E. 

X' = 74° 0' W. 

\g= 1°15 / E. 

X" = 72° 45' W. 



MIDDLE LATITUDE SAILING 339 

Solution II. By inspection. In Fig. 20 it is evident that if we enter 
the Traverse Table with the course and distance, the difference of latitude 
and departure may be found as in Plane Sailing, p. 329, Example 1. 
Corresponding to the course E.S.E. and distance 62, Table VIII, p. 83, 
gives the difference of latitude 23.7 and the departure 57.3 ; whence, the 
latitude in and middle latitude may be found as in Solution I. 

In Fig. 20 the difference of longitude AE is the distance corresponding 
to the course E and the departure AB. But the angle E = 90° — BAE ; 
that is, the angle E is the complement of the middle latitude. In this 
example the middle latitude is 40° 16' ; whence, the complement of the 
middle latitude is 90° - 40° 16' = 49° 44' = 50°, approximately. Corre- 
sponding to the course 50° and the departure 57.3 (the nearest is 57.5), 
Table IX, p. 131, gives the distance 75, which is the difference of longi- 
tude required ; whence, the longitude in may be found as in Solution I. 

Example 2. A ship sails from latitude 20° N., longitude 
160° W., until the difference of latitude is 1° 50' S. and the 
departure 440.2 miles W. Find by inspection the latitude in, 
the longitude in, the course, and the distance. 

U = 20° 0' N. L' = 20° 0' N. X' = 160° 0' W. 

L d = 1° 50' S. L" = 18° 10 / N. \ d = 7° 45' W . 

L" = 18° 10' N. 2 )38° 10' X" = 167° 45' W. 

i m =19° 5'N. 
co. L m = 70° 55' 
To find the longitude in. Since 440.2 exceeds the limits of the table, 
one third of this number is taken. 440.2 -f- 3 = 146.7. Corresponding 
to course 71° (= co. L m ) and departure 146.7, Table IX, p. 110, gives 
the distance 155. 3 x 155 = 465 = 7° 45', which is the difference of 
longitude ; whence, the longitude in is found to be 167° 45' W. 

To find the course and the distance. Since the departure 440.2 exceeds 
the difference of latitude 110 (= 1° 50'), the course and designation of the 
columns must be sought at the bottom of the page. 110 -=- 2 = 55, and 
440.2 -f 2 = 220. 1. Corresponding to the difference of latitude 55 and the 
departure 220.1 (the nearest are 54.9 and 220.3), Table IX, p. 105, 
gives the distance 227. 2 x 227 = 454. At the bottom of the page the 
course 76° is found. Hence, the course and the distance required are 
S. 76° W. and 454 miles. 

Example 3. A navigator wishes to sail from the Lizard, in 
latitude 49° 57' K, longitude 5° 14' W., to St. Mary's Island, 



340 



NAVIGATION 



in latitude 37° N. and longitude 25° 6' W. 
course and the distance. 



Find the true 



To find L d : 

X' = 49°57'N. 

£" = 37° Q'N. 

L d = 12° 57' S. 

= 777 miles. 

To find the course 
By [6], tan C 



To find L m : 
2/ = 49° 57' N. 
Z" = 37° / N. 

2 )86° 57 / 
X m = 43° 29' N. 



A„ COS i, 



X d = 1192 

L m = 43° 29' 
L d = 777 



logXd = 3.07628 
logcos£ m = 9.86068 
cologX rf = 7.10958-10 



log tan C = 10.04654 
.-. C = S. 48° 4' W. 



To find \ d : 

y = 5° 14 r W. 

X /r = 25° 6'W. 

\ d = 19° 52' W. 

= 1192 miles. 



To find the distance : 

By [2], D = L d secC 

log i d = 2.89042 

log sec C = 0.17505 

log D = 3.06547 

.-. B = 1163 miles. 



EXERCISE IV 





L' 


L" 


A' 


A" 


C 


D 


1 


25° 35' N. 


27° 28' N. 


60° 0'W. 


54°55'W. 


Required 


Required 


2 


32° 30' N. 


34° 10' N. 


25°24'W. 


29° 8'W. 


Required 


Required 


3 


39° 30' S. 


41° 0'S. 


74°20'E. 


70°12'E. 


Required 


Required 


4 


46° 24' S. 


Required 


178° 28' E. 


Required 


S.E. |E. 


278.0 


5 


20° 29' N. 


Required 


179° 10' W. 


Required 


W.byS.iS. 


333.0 


6 


0° 56' N. 


Required 


29°50'W. 


Required 


S. 47° E. 


168.0 


7 


42° 25' N. 


Required 


66° 14' W. 


Required 


S.E. by E. 


25.0 


8 


42° 8 r N. 


Required 


65°48'W. 


Required 


E.iS. 


126.0 


9 


41° 52' N. 


Required 


62°47'W. 


Required 


E.iS. 


161.0 


10 


41°38 / N. 


41°26'N. 


59° 16' W. 


Required 


E. by S. 


Required 


11 


41° 19' N. 


41°11'N. 


57°47'W. 


Required 


Required 


167.0 


12 


46° 28' N. 


45° 17' N. 


22° 18' W. 


19°39'W. 


Required 


Required 


13 


25° 30' S. 


28° 15' S. 


2°15'E. 


11°17'E. 


Required 


Required 


14 


33° 40' N. 


30°49'N. 


13°20'E. 


17°56'E. 


Required 


Required 


15 


19° 30' S. 


17°24'S. 


0°10'E. 


1°28'W. 


Required 


Required 



MIDDLE LATITUDE SAILING 341 

16. A ship sails from Boston lighthouse, in latitude 42° 20' 
N., longitude 71° 4' W., on a N.K.E. course, 184 miles. Find 
the latitude in and the longitude in. 

17. A ship sails from Cape May, in latitude 38° 56' X.. 
longitude 74° 57' W., on a S.S.E. course, 240 miles. Find 
the latitude in and the longitude in. 

18. A ship sails from Cape Cod light, in latitude 42° 2' N., 
longitude 70° 3' W., on an E. by X. compass course. 170 miles ; 
wind S.E. by S., leeway \ point, deviation 17f° E., variation 
11^° W. Find the latitude in and the longitude in. 

19. A ship sails from Cape Cod light, on a S.S.E. compass 
course, 140 miles ; deviation 5£° E., variation 11£° W. Find 
the latitude in and the longitude in. 

20. A ship sails from latitude 55° 1' X.. longitude 1° 25' W., 
on a S.W. compass course, 101 miles ; wind W.X.W., leeway 
1£ points, deviation 6° W., variation 24° 56' W. Find the 
latitude in and the longitude in. 

21. A ship sails from the Bermudas, in latitude 32° 18' X".. 
longitude 64° 50' W., on a W.S.W. compass course, 190 miles; 
deviation 1 point W., variation 1 point W. Find the latitude 
in and the longitude in. 

22. A ship sails from the Bermudas, on a W.X.W. com- 
pass course, 90 miles ; wind S.W., leeway 1 point, deviation 
1 point E., variation 1 point W. Find the latitude in and the 
longitude in. 

23. A navigator wishes to sail on a rhumb from the Ber- 
mudas to Cape Fear, in latitude 33° 52' X.. longitude 78° W. ; 
variation 10° W., deviation 7° W. Find the compass course 
and the distance. 

24. A ship from latitude 36° 32' X". sails between south and 
west until she has made 480 miles of departure and 9° 22' of 
difference of longitude. Find the latitude in, the course, and 
the distance. (Take L m = £(Z' + i") + 13'.) 



342 NAVIGATION 

SECTION XV 
MERCATOR'S SAILING 
Mercator's Chart 

Mercator's chart represents the surface of the earth as a 
rectangle. The meridians are represented by straight lines 
perpendicular to a straight line representing the equator, and 
the parallels are represented by straight lines parallel to the 
line representing the equator. On this chart any two merid- 
ians intercept equal distances on all the parallels; but on 
the earth the intercept on any parallel by two meridians is- 
less than the intercept on the equator. 

Hence, at no place except the equator does this chart cor- 
rectly represent the intercepts on the parallels, and the error 
increases as we approach the poles. 

In order to compensate for this error, the distance of each 
parallel from the equator is proportionally increased. 

These augmented latitudes aro called meridional parts, and 
are given for every minute of latitude in Table XI. 

On account of the spheroidal form of the earth, the develop- 
ment of the formula for the computation of this table is too 
difficult for presentation in an elementary work. 

The meridional difference of latitude is the expanded arc 
of the meridian between two parallels. It may be found b}^ 
taking the difference or the sum of the meridional parts corre- 
sponding to the two latitudes, according as the latitudes are 
of the same name or of opposite names. 

If the earth is regarded as a sphere, 

By [4], \ d =psecL. 

But on Mercator's chart, p = \ d ; that is, p is too great in 
the ratio of the secant of the latitude. Hence, the intercepts 
of the parallels on the meridians must be increased in the 



MERCATOR'S SAILING 



343 



^ .. 1 1 1 1 


— i 1 — i — i — | — i — l — i— r 


— l i i i 


i — i — 1 1 i — 


? 


■ 










■ 


- 


^ 


/ . 






■ 


- 


n 


f> 




B 


■ 


- 




mZ- 




■ 








- 


- 


c/ 








■ 


■ 


A 








- 


) i i i r 


1 1 i 1 


Jill 


i — r— f -i- 


\ — i — r r 


S 



55 



50 



35 



30 



25 20 



5 "West Long. 



Fig. 21 



344 NAVIGATION 

same ratio. But the intercepts thus augmented must be 
taken very small, otherwise a great amount of error will 
be introduced, because the triangles formed by the expanded 
latitude and the expanded departure cannot be regarded as 
similar plane triangles unless the distances are small. The 
augmented latitude for n degrees of latitude would be 
l'(sec 1' -f sec 2' + sec 3' H + sec n'). 

The construction and use of Mercator's chart will be under- 
stood by referring to Fig. 21,, which represents a chart extending 
from 30° to 60° north latitude, and from 0° to 25° west longitude. 

The horizontal line representing the parallel of 30° is first 
drawn and divided into equal parts. Through the points of 
division the meridians are drawn perpendicular to the parallel. 

In this case each division of the parallel represents 5° = 300'. 

From the Table of Meridional Parts, 

Meridional parts corresponding to a latitude of 30° = 1876.9' 
Meridional parts corresponding to a latitude of 35° = 2231. V 
Meridional difference of latitude for 5° = 354.2' 

Hence, the parallel of 35° is drawn at a distance of 354.2' 
above the parallel of 30°. 

Meridional parts corresponding to a latitude of 35° = 2231.1' 
Meridional parts corresponding to a latitude of 40° = 2607.9' 
Meridional difference of latitude for 5° = 376.8' 

Hence, the parallel of 40° is drawn at a distance of 376.8' 
above the parallel of 35° ; and so on. 

Let it be required to find from the chart the course and 
distance by rhumb line from A, in latitude 35° N., longitude 
20° W., to B, in latitude 50° K, longitude 5° W. 

A is at the intersection of the meridian of 20° and the par- 
allel of 35°; B is at the intersection of the meridian of 5° 
and the parallel of 50°. Draw AB. AB represents the 
rhumb line from A to B, for it makes the same angle C 
with all of the meridians. 



MERCATOR'S SAILING 



345 



The course C may be found by making one edge of a pair of 
parallel rulers coincide with AB and extending until the other 
edge passes through the centre of the compass-rose above. 
The rhumb line of the compass-rose which most nearly coin- 
cides with the edge of the ruler shows the course. 

The distance may be found by carrying M, the middle point 
of AB, over to the side, and measuring up and down a distance 
equal to one half of AB. The number of degrees between the 
extreme points reduced to miles is the distance required. 

It is evident that Mercator's chart shows relative positions 
accurately, but not relative distances. 

Let CA (Fig. 22) represent the ship's track. 
CB will represent the true difference of lati- 
tude, and BA the departure. Take CE equal 
to the meridional difference of latitude corre- 
sponding to CB, and through E draw ED 
parallel to BA to meet CA produced in D. 
Then, CD will represent the ship's track on 
Mercator's chart, and ED will represent the 
difference of longitude. fig. 22 

CE = Mer. L,, = meridional difference of latitude. 




CB = L, 



BA = p. 



ED = \ d . 



In the right triangle CED, 



or 





tan C = — , 




tan C - Xi 




tanL -Mer.L d ' 




.'. X d = Mer. L d x tan C. 


Also, by [3], 


tan C = f- ■ 




m % Mer. L d x p 

• • A-d — T 
^d 


Also, by [2], 


D = L d sec C; .'.I 



[8] 



[9] 



n cos c. 



346 



NAVIGATION 



To find the difference of longitude made when the ship 
sails on an oblique course there are two distinct methods : 
Middle Latitude Sailing and Mercator's Sailing. 

When to use Middle Latitude and Mercator's Sailings 



Either Middle Latitude or Mercator's Sailing may be used 
when the distance is small (such as an ordinary day's run) ; 
but when the distance is great, Mercator's Sailing should be 
employed if the course is less than about 45°; and Middle 
Latitude Sailing should be employed if the course exceeds 
about 45°. 

This distinction is of special importance in high latitudes. 

Example 1. A ship sails E.N.E., 200 miles, from latitude 
40° 45' K, longitude 74° W. Find the latitude in and the 
longitude in. 



Solution I. By computath 
To find the latitude in : 
By [2], L d = D cos C. 



C = 67° 30' 



log D = 2.30103 
log cos C = 9.58284 
log i d = 1.88387 
. L d = 76.5 miles 
= 1 17' N. 
2/ = 40° 45' N . 
L" = 42° 2' N. 



To find the longitude in : 
U = 40° 45' N., Mer. Parts = 2666.8 
2/' = 42° 2'N., Mer. Parts = 2769.0 
Mer. L d = 102.2 

B y [ 8 ]> K = Mer. L d x tan C. 
log Mer. L d - 2.00945 
log tan C = 10.38278 
logXrf= 2.39223 
;.\ d = 246.7 miles 
= 4° 7'E. 
X' = 74° / W . 
.-. X" = 69° 53' W. 

Solution II. By inspection. E.N.E. = 6 points. Corresponding to 
a course of 6 points and a distance of 200, the true difference of latitude 
is found in Table VIII to be 76.5 ; whence, the latitude in and the merid- 
ional difference of latitude are found as in Solution I. Corresponding to 
the course 6 points, and the meridional difference of latitude 102.2 as a 
difference of latitude, the same table gives the departure 246.7, which is 



MERCATOR'S SAILING 



347 



equal to the difference of longitude required 
may be found. 



whence, the longitude in 



Example 2. Required the course and the distance from 
Halifax, in latitude 44° 40' N. and longitude 63° 35' W., to 
Ireland island, in latitude 32° 19' K and longitude 64° 49' W. 





L' = 44° 40" N. 


Mer. Parts : 


= 2985.6 


X' =63° 


35' W. 


L" - 32° 19' N. 


Mer. Parts = 2038.0 X" = 04° 40' W. 


L d =i2 o 2rs. 


Mer. L d = 947.0 X rf = 1° 14' W. 


= 741 miles. 


= 74 miles. 


To find the course : 




By [8], 


tan C = 


A rf 




Mer. L d 


\d = 74, 


logX d = 1.86923 


Mer. L d = 947 


colog Mer. L d = 7.02365 - 10 




log tan (7 = 8.89288 




.-. C = S. 4° 28' W, 


To find the distance : 




By [2], 


D = L d sec C. 


L d = 741 


, log L d = 2.86982 


C = 4°! 


28'. log sec C = 0.00132 




logD = 2.87114 




.-. D = 743.3 miles. 




EXERCISE V 




V 


L" 


A' 


A" 


C D 


1 


38° 14' N. 


39° 51' N. 


2° 7'E. 4°18'E. 


Required | Required 


2 


49° 53' N. 


48° 28' N. 


6° 19' W. 


5° 3'W. 


Required Required 


3 


64° 30' N. 


60° 40' N. 


4° 20' W. 


0° 10' E. 


Required Required 


4 


54° 54' S. 


34° 22' S. 


60° 28' W. 


18° 24' W. 


Required Required 


5 


17° O'N. 


20° O'N. 


180° 0'E. 


177° 0'E. 


Required 


Required 


6 


45° 15' N. 


Required 


35° 26' W. 


Required 


N. 49° E. 


175 


7 


55° l'N. 


Required 


1°25'E. 


Required 


N. 10° E. 


246 


8 


50° 48' N 


Required 


9° 10' W. 


Required 


S. 41° W. 


275 


9 


37° O'N. 


51° 18' N. 


48° 20' W. 


Required 


Required 


1027 


10 


51° 15' N. 


37° 5'N. 


9° 50' W. 


Required 


S.W.byS. 


Required 



348 NAVIGATION 



11. Required the course and the distance from Toulon to 
Valencia, by Mercator's Sailing : 

27' N. 
19' W. 



_ 1 fL=±3° 8'N. ^ T . . CL = 39 c 

T ° ul0n U= 5°56'E. ^ alenCia U= 0< 



12. Required the compass course and the distance from Cape 
East, New Zealand, to San Francisco ; variation 14° 20' E. and 
deviation 5° 40' E. : 

L= 37°48'N. 
X = 122° 24' W. 



C L = 37° 50' S 
Cape East \ x = 17 go 36 , E San Francisco 



13. Required the course and the distance from Cape Lopatka 

to Callao : 

4'S. 

14' W. 



Cf— 51° 9'N f L = 19° 

Cape Lopatka j ^ = 1560 ^ R Callao ( ^ = ^. 



14. A ship from latitude 20°40'N. sails N.E. by NT. until 
she is in latitude 27°16'N. Required the distance and 
the difference of longitude. 

15. A ship from Cape Clear, in latitude 51°26'N. and 
longitude 9°29'W., sails S.W. by S. until the distance run 
is 1022 miles. Find the latitude in and the longitude in by 
Mercator's and Middle Latitude Sailings. Which method is 
preferable ? 

SECTION XVI 

TRAVERSE SAILING 

When a ship reaches her destination by sailing on different 
courses the difference of latitude may be found by computing 
the change in latitude for each course separately, as in Plane 
Sailing, and taking the algebraic sum ; or, more conveniently 
(but less accurately), by writing the north latitudes in one 
column and the south latitudes in another, finding the sum of 
each column and taking the arithmetical difference, giving it 



TRAVERSE SAILING 



340 



the name (N. or S.) of the greater sum. In like manner 
the total departure may be 
found approximately. 

Evidently the Traverse 
Table may be used and the 
work arranged as in Rec- 
tangular Surveying. 

When the difference of 
latitude and the departure 
have been found, the differ- 
ence of longitude may be 
determined by Middle Lati- 
tude or Mercator's Sailings. 

Example 1. A ship in latitude 40° N. and longitude 67° 4' 
W. sails N.W. 60 miles, then N. by W. 52 miles, then W.S.W. 
83 miles. Find the latitude in and the longitude in. 




Fig. 23 



c 


D 


N 


S 


E 


W 


N.W. 

N. by W. 
W.S.W. 


4 points 
1 point 
6 points 


60 
52 

83 


42.4 
51.0 


31.8 




42.4 

10.1 

76.7 


Hence, La = 61.6 miles N. 
= 1° 2' N. 


93.4 
31.8 


31.8 





129.2 
0. 


1 


\ = 129.2 W 




61.6 


129.2 



L" 




L m 


\" 


U =40°0 / N. 


L' 


= 40° 0' N. 


V =67° 4' AY. 


L d = 1° 2' N. 


L" 


= 41° 2'N. 


X ( , = 2°51'W. 


£"=41° 2' N. 




2)81° 2' 


X" = 69° 55' W. 



L m = 40° 31' N. 
co. L m = 49° 29' 

' The difference of longitude may be found by computation, using for- 
mula [5], or by inspection, as in Sect. XV, p. 347, Example 2. 49° is found 
at the bottom of p. 132, Table IX, and opposite the departure 129.1 is 
distance 171, which is the difference of longitude required. 171' = 2° 51'. 



350 



NAVIGATION 



To plot the Courses 

Let C (Fig. 24) be the centre of a circle representing the 
card of the compass. Draw CA, N.W. from C, equal to 60, 
to any convenient scale. Make an edge of a pair of parallel 
rulers coincide with the N. by W. rhumb of the compass 
diagram, extend until the other edge embraces A, and draw 
AB equal to 52. Make an edge of the ruler coincide with the 
W.S.W. rhumb of the compass diagram, extend until the other 
edge embraces B, and draw BD equal to 83. The point D 
represents the position of the vessel. 




Fig. 24 



Upon CN produced let fall the perpendicular DF. Then, 
CF represents the total difference of latitude, CD the distance 
made good, and DCF the course made good. 

The departure may be regarded as measured on the parallel 
i(L' + L"). 

Example 2. The U.S. steamer Swatara left latitude 37° 25' 
K and longitude 3° 14' E. at noon, Feb. 25, 1880, and during 
the next 24 hours sailed on the (corrected) courses in the 



TRAVERSE SAILING 



351 



following table. Find the latitude in and the longitude in by 
account, at noon Feb. 26. 



c 




D 


X 


S 


E 


W 


X.E. byE. |E. 


5£ points 


2.0 


0.9 




1.8 




E. by N. 


7 points 


28.0 


5.5 




27.5 




E.fS. 


1\ points 


3.0 




0.3 


3.0 




E.S.E. iE. 


Q\ points 


2.8 




0.8 


2.7 




S.S.E. £E. 


2-J- points 


3.4 




3.0 


1.6 




X. by W. i W. 


1£ points 


4.2 


4.1 






1.0 


N. irE. 


£ point 


4.4 


4.4 




0.2 




N. byE.±E. 


1± points 


10.G 


10.3 




2.6 




N.£E. 


| point 


11.4 


11.4 




1.1 




N. i W. 


£ point 


18.6 


18.5 






1.8 


N. by W. 


1 point 


34.8 


34.1 






6.8 


S. byE. 4E. 


\\ points 


5.0 




4.8 


1.5 










89.2 


8.9 


42.0 


9.6 


Hence, La = 


BO. 8 = 1°20' 


N., 


8.9 




9.6 




and p = 


32.4 E. 






















80.3 




32.4 





L' = 37° 25' N. 
L d = 1° 20' N . 
L" = 38° 45' N. 



37° 25' X. 

38° 45' N. 



2 )70° 10' 
Z,,„ = 38° 5'N. 
co. L m = 51° 55' 



X' = 3°14'E. 

X, = 0°41'E . 
X" = 3° 55' E. 



EXERCISE VI 

Required the difference of latitude and the departure : 



S.S.W. 
S.W. by S. 

N.E. 



48 
36 
24 




c 


D 


S.S.W. JW. 
S.S.W.{W. 
S. by W. i W. 


43 
39 
27 



352 



NAVIGATION 



4 




C 


D 


N. 25° W. 


16.4 


N. 8°E. 


7.8 


N. 19° E. 


13.7 


N. 76° E. 


39.6 



C 


D 


WJ.¥.|W. 
N.N.E. |E. 
N. by E. f E. 
S.S.W.iW. 


21 
9 
9 

30 



C 


D 


S. 83° W. 


23.0 


S. 48° E. 


25.2 


N. 48° W. 


27.1 


N. 36° W. 


21.0 



C 


D 


S. 17° E. 


48 


S. 45° W. 


19 


N. 36° W. 


18 


N. 41° W. 


50 


E. 


36 



8 




C 


D 


N.N.E. 


31 


E.N.E. 


35 


E. by S. 


36 


S.S.E. 


51 


S. by E. 


60 



C 


D 


S. 44° E. 


69 


S. 85° E. 


68 


S. 27° E. 


25 


N. 37° W. 


5 \ 


N. 20° W. 


» 



SECTION XVII 



THE DAY'S WORK 



When a ship is starting out upon a voyage, just before she 
passes out of sight of land the proper officer notes the bear- 
ing, and estimates the distance from the ship, of some object 
on shore whose latitude and longitude are known. This oper- 
ation is called taking the departure. After this has been done, 
a record is kept on the log board of the courses and the dis- 
tances run. From the log board the record is transferred 
to the log book, in which it is arranged as in the tables on 
pages 354 and 357. 

The column headed H contains the hours from noon to 
noon. The column headed K contains the knots per hour the 
ship has sailed ; and the next column contains the tenths of 
knots. The column headed Winds shows the direction of the 



THE DAY'S WORK 353 

wind ; the column headed Leeway gives the leeway in points ; 
the column headed Deviation gives the deviation in degrees ; 
and the last column contains a record of miscellaneous 
events. 

From the record of the log book the position of the ship is 
computed by Traverse Sailing each day at noon, or at any 
other hour when the position is desired, in the manner 
described in the examples on pages 356 and 358. 

In computing the day's work, the first course of the traverse 
is found by reversing the departure course. For if, as in 
Example 1, the bearing of the object from the ship is N.E. £ E. 
15 miles, then it follows that the bearing of the ship //•<>?», the 
object on shore is S.W. \ W. 15 miles. 

This method of finding the place of a ship is subject to many 
errors. It is impossible to determine the course accurately ; 
the deviation is almost constantly changing ; the effect of tides 
and currents can be estimated only approximately ; the log 
gives the distance roughly ; the speed varies during the hour ; 
and so on. 

Hence, whenever it is possible to do so the results obtained 
by dead reckoning should be verified by celestial observation. 
(See Chapter III.) The results obtained by the latter method 
should be given the preference. 

A number of the following exercises are taken from recent 
examination papers of the British Marine Board ; others are 
actual transcripts from the log books of vessels. 

The log book used in the U.S. navy contains, in addition to 
the usual data, columns in which are recorded the force of the 
wind, the height of the barometer, the temperature of the air 
by dry and wet bulb thermometers, the surface temperature of 
the water, the state of the weather, the forms of clouds, the 
proportion of clear sky, the state of the sea, a record of the 
sail the vessel is under at the end of the watch ; also, a record 
of the miscellaneous events of the day. 



354 

Example 1. 



NAVIGATION 



H 


Courses 


K 

10 


8 


Winds 


Leeway 


Deviation 


Remarks, etc. 


1 


W.S.W. 


N. W. 


ipt. 


11° W. 


The departure was 
taken as follows : A 


2 




11 


4 








point of land in lati- 
tude 37° 3' N., longi- 


3 




11 


4 








tude 9° 0' W., bear- 
ing by compass N E 
i E., dist. 15 miles. 


4 




11 


4 






















Ship's head W.S.W. 


5 


N.W. | N. 


12 


2 


w.s.w. 


fpt. 


17° W. 


Deviation as per log 


6 




12 


3 










7 




12 


3 










8 




12 


2 










9 


N.N.W. 


9 


6 


w. 


£ pt. 


11° W. 




10 




9 


5 










11 




9 


5 










12 




9 


4 








Variation 22° 30' W. 


1 


N.W.byW. 


7 


8 


S.W.byW. 


1± pts. 


20° W. 




2 




7 


6 










3 




7 


4 










4 




8 


2 








A current set the ship 
S.W. by W. | W. 


5 


S.W. |S. 


9 


3 


S.S.E. 


l pt. 


6°W. 


(correct magnetic) 8 
miles from the time 


6 




8 


7 








the departure was 










taken to the end of 


7 




9 


3 








the day 


8 




8 


7 










9 


W.iS. 


10 


3 


S. by W. 


ipt. 


13° W. 




10 




10 


2 










11 




10 


2 










12 




10 


3 











Departure course. 

The opposite point to N.E. J E. is 
S.W. i W. The ship's head being 
W.S.W., the deviation is the same 
as for the first course, 11° W. 

S.W. I W. =4ipts. R. of S. 

or 47° 49' R. of S. 

Dev. 11° 0'W. 

Var. 22° 30' 

True course = 14° 19' R. of S. 



YW.f 



33° 30' L. 



Hence, the first course and the 
distance of the traverse are 
S. 14° W. 15 miles. 
First course. 

= pts. R, of S. 
= jpt. L. 



W.S.W. 

Leeway 



5£ pts. R. of S. 
61° 53' R. of S. 



or 
Dev. 11° 0'W. 
Var. 22° 30' W. 
True course = 28° 23' R. of S. 



-:} 



33° 30 r L. 



THE DAY'S WORK 



355 



The distance (45 miles) for this 
course is found by adding the dis- 
tances sailed each hour up to 5 
o'clock, when the course was 
changed. 

Hence, the second course and the 
distance of the traverse are 
S. 28° W. 45 miles. 
Second course. 
N.W.|N. = H pts. L. of N. 

Leeway = j pt. R. 

3i pts. L. of X. 
36° 34' L. of X. 



39° 30' L. 



or 
Dev. 17° 0' W. 
Var. 22° 30' W. 
True course = 76° 4' L. of X. 

The distance (49 miles) is obtained 
by adding the hourly distances from 
5 o'clock to 9 o'clock. 

Hence, the third course and the 
distance of the traverse are 
N. 76° W. 49 miles. 

Third course. 
N.N.W. = 2 pts. L. of X. 

Leeway = f pt. R. 

U pts. L. of N. 
14° 4' L. of N. 



33° 30' L. 



or 
Dev. 11° 0' W. 
Var. 22° 30' W. 
True course = 47° 34' L. of X. 

The distance (38 miles) is obtained 
by adding the hourly distances from 
9 o'clock to 1 o'clock. 

Hence, the fourth course and the 
distance of the traverse are 
N. 48° W. 38 miles. 

Fourth course. 
N.W. by W. = 5 pts. L. of X. 

Leeway = lj pts. R. 

3f pts. L. of X. 



42 11' L. of X. 
Y W. J' 42 °° L - 



or 
Dev. 20° (TV 
Var. 22° 30' 
True course = 84° 41' L. of X. 

Hence, the fifth course and the 
distance of the traverse are 
N. 85° W. 31 miles. 

Fifth course. 
S.W. £S. =3* pts. R. of S. 

= 1 pt. R. 
4£ pts. R. of S. 
50° 38' R. of S. 



Leeway 



or 
Dev. 6° 0'TV 



.22°30'W. I" 28 3 ° K 



Var, 

True course = 22 : 8' H. of S. 

Hence, the sixth course and the 
distance of the traverse are 
S. 22° W. 36 miles. 



Sixth course. 
W. i S. 
Leeway 



= 7£ pts. R. of S. 
= ipt R. 



8 pts. R. of S. 
or 90° 0' R. of S. 

Dev. 13° 0-W.i 

Var. 22= 30' W. j ~ _ *" 
1 True course = 54° 30' R. of S. 

Hence, the seventh course and the 
I distance of the traverse are 
S. 55° W. 41 miles. 

Current course. 
S.W. by W. £ W. = 5| pts. R. of S. 

or 64° 41' R. of S. 

Variation = 22° 30' L. 

True course = 42° 11' R. of S. 

Hence, the eighth course and the 
distance of the traverse are 
S. 42° TV. 8 miles. 



356 



NAVIGATION 



The Traverse 



1 


Courses 


DlST. 


N 


s 


E 


w 


S. 14° W. 


15 




14.6 




3.6 


2 


S. 28° W. 


45 




39.7 




21.1 


3 


N. 76° W. 


49 


11.9 






47.5 


4 


N. 48° W. 


38 


25.4 






28.2 


5 


N. 85° W. 


31 


2.7 






30.9 


6 


S. 22° W. 


36 




33.4 




13.5 


7 


S. 55° W. 


41 




23.5 




33.6 


8 


S. 42° W. 


8 




5.9 




5.4 




L d = 77.1 miles 
= 1 17' S. 


40.0 


117.1 
40.0 




183.8 


p = 183.8 ^ 


r. 


77.1 



To find the latitude in : 

JJ =37° 3'N. 

L d = i°irs. 

L" = 35° 46' N. 
Also, L m = 36° 25' N. 

To find the direct course : 
By [3], tan C = f- • 

logp = 2.26435 
colog L d = 8.11295-10 
log tan C= 10.37730 

.-. C = S. 67° 15' W. 

To find the direct distance : 
P 



By[l], d = 



sin C 



logp = 2.26435 
colog sin C = 0.03517 
log D = 2.29952 
.-. B = 199 miles. 

To find the difference of longitude 

% [ 5 ]> \ d =psecL m . 

logp = 2.26435 

log sec L m = 0.09435 

log \ d = 2.35870 

.-. \ d = 228' = 3° 48' W. 

To find the longitude in : 

X' = 9° 0'W. 

\ d = 3° 48' W . 
\" = 12° 48' W. 



THE DAY'S WORK 



357 



Example 2. 



H 

1 
2 


Courses 
S.W. 


K 

6 
6 



3 


Winds 


Leeway 


Deviation 


Remarks 


w.x.w. 


1 pt. 





Latitude from 

•2H C 30' S. 
I.onsritude from 

10° 15' E. 


3 




6 


4 










4 




G 













5 




5 


3 













S. by W. 


6 





W. by S. 


lpt. 






7 




5 


1 










8 




5 


4 










9 




5 


2 










10 




5 


3 








Variation U pts. W. 


11 




5 













12 


S.S.W. 


5 


2 


W. 


1 pt. 






1 




5 


2 










2 




5 













3 
4 
5 




4 

5 

5 


6 










A current set S. (cor- 
rect magnetic) 2J 
miles per hour 
from noon to noon 


6 


S.E. by S. 


5 


2 


S.W. by S. 


1 pt. 






7 




5 


4 










8 







5 










9 




6 













10 




6 













11 




5 


4 










12 




5 


5 









First course. 

S.W. 

Leeway 

Variation 
True course 



pts. R. 
pt. L. 



of S. 



3 pts. R. 
1£ pts. L. 



of S. 



= H pts. R. of S. 
= S. 1£ pts. W. 



Second course. 

S. by W. 
Leeway 

Variation 
True course 



= 1 
= 1 



pts. 
1-J- pts. 



of S. 



H pts. L. of S. 
S. H Pts. E. 



358 



NAVIGATION 



= 2 
= 1 



pts. R 
pt. L. 



of S. 



Third course. 
S.S.W. 
Leeway 

Variation 
True course = i pt. L. of S. 
= S, i pt. E. 
Current course. S. 

Variation 



1 pt. R. of S. 

=: 1£ ptS. L. 



Fourth course. 
S.E. by S. 
Leeway 

Variation 



pts. L, 
pt. L. 



of S. 



4 pts. L. of S. 
H Pts. L. 



True course = 5£ pts. L. of S. 
= S. 5£ pts. E. 
pts. L. of S. 
H pts. L. 



True course = 1£ pts. L. of S. 
= S. H pts. E. 



The Traverse 





Courses 


DlST. 


N 


S 


E 


W 


1 

2 
3 
4 
5 


S. H points W. 
S. 1£ points E. 
S. i point E. 
S. 5-$- points E. 
S. 1£ points E. 


30 
32 
30 
39 
60 




28.7 
30.6 
29.9 
18.4 
57.4 


9.3 

2.9 

34.4 

17.4 


8.7 


L d - 165 miles = 2°45 , S. 
L' = 28° 30' S. 




165.0 


64.0 

8.7 


8.7 




L" = 31° 


15' S. 


55.3 



~L' = 28° 30 7 S. Mer. Parts = 1774.3 

L" = 31° 15' S. Mer. Parts = 1963.6 

Mer. L d = 189.3 
To find the difference of longitude : 

_ Mer. L d x p 

log Mer. L d = 2.27715 

logp = 1.74273 

colog L d = 7.78252 - 10 

log\d = 1.80240 

.-. X d = 63.4 = 1° 3 7 E. 



By [9], 



Mer. L d = 189.3, 

p= 55.3, 

L d = 165.0. 



X = 10° 15' E. 
\ d = 1° 3'E. 
\" = 11° 18' E. 



Note. Find the course by [3] to determine whether to use Middle 
Latitude or Mercator's Sailing. 

The distance made good may be found by [2]. 



THE DAY'S WORK 



359 



EXERCISE VH 

Find the latitude in and the longitude in from the follow- 
ing log-book records : 

1 



H 


Courses 


K 


T 


WINDS 


Leeway 


Deviation 


Remarks 


1 


N.N.E. 


6 


3 


w. 





f) Latitude from 
46° 2s- V 


2 




6 


2 








Longitude from 

.'■-' : u w. 


3 




6 


5 










4 




6 


4 










5 




6 













6 


E.X.E. 


6 


1 


N.W. 








7 




6 


6 










8 




5 


8 










9 




5 


6 










10 




5 


4 










11 




5 


5 










12 


E. by S. 


5 


3 


N. 






Variation 


1 




5 


9 










2 




6 


2 










3 




6 













4 




6 


3 










5 




6 


4 










6 

7 


S.S.E. 


7 
6 




8 


N. 






The tide set S. by 
E. 2 J miles per 
hour during the 
24 hours 


8 




7 


3 










9 




7 


5 










10 




7 


1 








" 


11 




7 


9 










12 




7 


o 
o 











360 



NAVIGATION 



H 

1 

2 


Courses 


K 


T 


Winds 


Leeway 


Deviation 


Remarks 


S. by W. 


6 
6 


2 



W. 








Latitude from 
33° 40' N. 

Longitude from 
16° 20' W. 


3 




6 


3 










4 




7 













5 




7 


2 










6 




7 


3 










7 


S.W. by S. 


7 


2 


W. by N. 


1 pt. 






8 




7 


2 










9 




7 


4 










10 




7 


6 










11 




7 


4 










12 




8 


1 








Variation 12° 20 "W. 


1 




8 













2 




8 


5 










3 




8 


2 










4 


S.W. by W. 


7 


5 


N.W. 









5 




7 


3 










6 

7 




6 
6 


6 
4 








A current set W.S. 
W. (correct mag- 
netic) 1J miles 
per hour from 
noon to noon 


8 




6 













9 




6 


2 










10 




6 


1 










11 




6 


3 










12 




6 


1 











THE DAY'S WORK 



361 



H 


Courses 


K 


T 


Winds 


Leeway 


Deviation 


Remarks 


1 

2 


N. by E. 


6 

6 


7 
2 


E. by N. 


lpt. 





Latitude from 

l'J° 30' S. 

Longitude from 

0° 10' E. 


3 




6 


4 










4 




6 


3 










5 




6 


1 










6 




6 













7 


N. 


5 


8 


E.N.E. 


1 pt. 






8 




5 


4 










9 




5 













10 




5 


3 










11 




5 


6 










12 




5 


9 








Variation 13° 30 W. 


1 




5 


7 










2 


N.N.W. 


6 


4 


N.E. 


1 pt. 






3 




6 


8 










4 




7 













5 




7 


3 










6 

7 




7 
7 


6 
5 








The current set W. 
N.W. (correct mag- 
netic) J mile per 
hour from noon to 
noon 


8 




7 













9 




7 


2 










10 




7 


4 










11 




6 


3 










12 




6 














362 



NAVIGATION 



H 


Courses 


K 


T 


Winds 


Leeway 


Deviation 


Remarks 


1 
2 
3 
4 


S.E. by E. 


13 
13 
12 
13 


2 
3 
5 


K. 





11° E. 


A point of land, lati- 
tude 47° 31' N., 
longitude 52° 33' 
W., bearing by 
compass W.S.W., 
distant 18 miles. 
Ship's head S.E. 
by E. Deviation 
as per log 


5 


S.E. 


11 




E.N.E. 


ipt. 


9°E. 




6 




10 


5 










7 




10 


4 










8 




11 


1 










9 


E. by N. 


8 


2 


S.E. by S. 


1 pt. 


17° E. 




10 




9 


4 










11 




9 


4 










12 




9 












1 


E.N.E. 


6 


8 


S.E. 


H pts. 


15° E. 


Variation 28° W. 


2 




6 


7 










3 




6 


5 










4 




7 












5 
6 

7 


S.S.E. 


5 

5 
6 


8 
8 
4 


E. 


2 pts. 


7°E. 


A current set the 
ship (correct mag- 
netic) S. by E. 12 
miles from the 
time the depar- 
ture was taken to 
the end of the day 


8 




6 












9 


S.E. by S. 


7 




E. by N. 


li pts. 


8°E. 




10 




7 


3 










11 




7 


4 










12 




7 


3 











THE DAY'S WORK 



363 



H 


Courses 


K 


T 


Winds 


Leeway 


Deviation 


Kemakks 


2 

o 
O 

4 


S.S.E. 


5 
5 
5 
4 


6 
6 


8 


E. 


2± pts. 


3° E. 


A point of land in 
latitude >'>1 : X., 
longitude 150° 
E.. bearing by 
compass W. hv 
S. J S., distant 
17 miles. Ship's 
head S.S.E. De- 
viation as per 
log 


5 


s.s.w.±w. 


4 


7 


W. 


2f pts. 


4° W. 




6 




4 


8 










7 




5 


2 










8 




5 


3 










9 


W.S.W. 


5 





s. 


2£pts. 


9° W. 




10 




6 













11 




6 


5 










12 




6 


5 










1 


W. } N. 


6 


6 


X. by E. 





11° W. 


Variation 31° E. 


2 




7 













3 




6 


4 










4 




6 













5 


E. 


4 


6 


S.S.E. 


2| pts. 


10° E. 




6 

7 
8 




5 
4 
4 



8 
6 








A current set the 
ship (correct mag- 
netic ) X.X.E. 21 
miles from the 
time the depar- 
ture was taken to 
the end of the day 


9 


E.S.E. 


4 





S. by W. 





9°E. 




10 




4 


5 










11 




4 


5 










12 




5 














364 



NAVIGATION 



H 


Courses 


K 


T 


Winds 


Leeway 


Deviation 


Remarks 


1 
2 
3 
4 


S.W. 1 W. 


14 
14 
14 
14 


5 

2 
6 

7 


S.E. 





6°W. 


A point, latitude 50 c 
12' S., longitude 
179° 40' W., bear- 
ing by compass N. 
i W., distant 19 
miles. Ship's 
head S.W. i W. 
Deyiation as per 
log 


5 


N. |E. 


4 





E.N.E. 


3± pts. 


3°E. 




6 




o 
o 


6 










7 




3 


6 










8 




3 


8 










9 


S. byE. i-E. 


2 


4 


S.W. iW. 


2f pts. 


6°E. 




10 




2 


3 










11 




2 


3 










12 




2 













1 


W. by S. 


12 


2 


S. by W. 


ipt. 


14° W. 


Variation 14° E. 


2 




12 


4 










3 




12 


6 










4 




12 


8 










5 
6 

7 
8 


E.N.E. 


3 

2 
3 
3 



3 
4 
3 


S.E. 


2£ pts. 


19° E. 


A current set the 
ship (correct mag- 
netic) S.W. J W. 
42 miles from the 
time the depar- 
ture was taken to 
the end of the 
day 


9 


S.S.W.|W. 


5 


6 


S.E. 


If pts. 


4°W. 




10 




5 


7 










11 




5 


3 










12 




5 


4 











THE DAY'S WORK 



365 



H 


Courses 


K 


T 


Winds 


Leeway 


Deviatiox 


Remarks 


1 
2 














.S™ 


3 
















4 
















5 
















6 
















7 
















8 


S.E. 


1 


5 







i pt. E. 




9 




6 


4 










10 




6 


6 










11 




6 


8 










12 




6 


5 










1 


E.S.E. ±E. 


7 


2 










Variation 1} pts. W. 


2 




7 


2 










3 




8 


2 










4 




8 


6 










5 




8 













6 




7 


6 










7 




7 


6 










8 




7 


6 










9 




7 


5 










10 




5 


'7 










11 


E. 


4 


2 







I pt. E. 




12 




4 


5 











366 NAVIGATION 

SECTION XVIII 
GREAT CIRCLE SAILING 

It may be proven {Geometry, § 816) that the shortest distance 
between two places on the surface of the earth is the less of 
the two arcs of a great circle lying between these places. The 
rhumb line is the arc of a great circle only when it coincides 
with the equator or a meridian. 

The advantages of great circle sailing are : 

(1) The distance is less than by any other route. 

(2) The great circle track gives the real direction of the 
desired port, and this frequently determines which tack the 
ship should be put on. 




Fig. 25 Fig. 26 

To sail on a great circle would require a constant change of 
course. This difficulty is practically overcome by sailing on 
rhumb lines from point to point of the great circle. 

The course of the vessel may be said to describe part of the 
perimeter of a polygon with a large number of sides within 
the arc of a great circle. 

Definition. The vertex of a great circle track is the point 
nearest the pole and therefore farthest from the equator, and 
touches the highest latitude. Hence, the vertex is the foot 
of the perpendicular let fall upon the great circle from the 



GREAT CIRCLE SAILING 



:367 



pole, because the perpendicular measures the shortest dis- 
tance. When the two places are widely separated, the vertex 
generally lies between the two places, but it by no means 
necessarily does lie between the two places. It will be found 
that if the two angles C and C (Figs. 25 and 26) are both 
greater than 90° or both less than 90°, the vertex will fall 
between the two places ; otherwise it w r ill not. 

The best practical method of finding the courses, distance, 
etc., of the great circle track is by the use of Godfray's Course 
and Distance Diagram, which affords a simple graphic solution. 

The solution by Spherical Trigonometry is given here. 



liiilllljiiiiiiiillllllllilliilii 





Fig. 27 



Fig. 2* 



Let CC (Fig. 28) represent the great circle track between 
C and C, P the pole of the equator, and PC and PC 
meridians. 

C = the first course from C. 
C = the first course from C. 

P = X d = the difference of longitude of C and C. 
PC =c' = the polar distance of C = 90° ip latitude of C. 
PC = c = the polar distance of C = 90° zp latitude of C. 
CC' = D = the length of the great circle track. 
In Navigation two sides of the spherical triangle are the two 
meridians passing through the places between which the ship's 



368 



NAVIGATION 



track lies, and the third side is the ship's track. The lengths 
of the two sides formed by the meridians are always known 
from the latitudes of the places to and from; one of the 
angles, that at the pole, is found from the difference of longi- 
tude between the two places. Hence, it is evident we have to 
solve a spherical triangle, given two sides and the included angle. 
The solution consists in determining the following elements : 

(1) The first and last courses. 

(2) The distance. 

(3) The position (latitude and longitude) of the vertex. 

(4) A succession of points on the great circle. 

To find the First Course 



In the triangle CPC' (Fig. 30) by Napier's Analogies 
(Spherical Trigonometry, Sect. LVI, p. 164) : 



tani(C + C') = 



tani(C 



C') = 



cos \ (c — c') 
cos \ (c + c') 
sin i (c — c r ) 
sini(c + c') 



C0ti\ d , 



cot 



C=KC + C') + f(C-G'), 

C' = i(C + C')-i(C-C). 



[10] 



To find the Distance 



By Gauss's Equations (Spherical Trigonometry, Sect. LVI, 
164): 



cos \ D = 



cosi(c + c') 



sin 



i** 



[11] 



cosi(C-i-C') 

(See also Spherical Trigonometry, Sect. LXV, p. 182.) 
The position of the vertex must be known to enable us to 

calculate the course and the distance from one point to another 

on the great circle. 



GREAT CIRCLE SAILING 



369 



To find the Latitude of the Vertex 

Let PV (Fig. 30) be the perpendicular from P upon the 
arc of a great circle CC. Then V is the vertex of CC. 
In the right triangle CVP, 

sin PV = cos (co. c') cos (co. C). 
Hence, sin PV = sin c' sin C, "1 

Latitude of V = 90° - PV. J ^ J 

(See Spherical Trigonometry, Sect. LI, p. 147.) 

To find the Longitude of the Vertex from C 

We use the known elements c' and C to find CPV. 
In the right triangle CVP (Fig. 30), 

sin (co. c')= tan (co. C) tan (co. CP V). 
.'.cose' = cot Ccot CPV. 
Hence, cot CPV = cos c' tan C. [13] 





Fig. 29 



Fig. 30 



By combining this result with the longitude of C, the longi- 
tude of V may be found. 

It is evident that if a series of points is laid down on the 
great circle track and the navigator sails the ship from point 



370 NAVIGATION 

to point by Mercator's or Middle Latitude Sailing, lie will 
shorten his route very considerably. It is easy to find a suc- 
cession of points, beginning at the vertex, since the angle at 
the vertex is a right angle. Hence, knowing the latitude and 
the longitude of different points on the track the navigator 
may sail from one point to another. 

To find the Latitude of Intermediate Points of Given 
Longitude, in order to plot the Great Circle 
Track on the Chart 

Let m (Fig. 31) be a point whose longitude from V is 

p assumed. 

/j^T^^ In the right triangle PVm, 

/ ( \\ \ \ sm {go. mP V)= tan P V tan (co. Pm) . 

/ c m^^A \ . • . cos raP F = tan P F cot Pm. 

Hence, 



cot Pm = cos mPV cot PV, 



} [14] 



\ 7 Latitude of m = ± (90° — Pm), 

fig 31 -*- accor( iing as m is above or below the 

equator. 

Points m, etc., are usually assumed on each side of V, differ- 
ing from V and from each other 5° or 10° in longitude. In 
this manner the latitude and longitude of as many points as 
we please between C and C may be determined. The course 
and distance from point to point by rhumb line may then be 
found by Middle Latitude or Mercator's Sailings. 

When either of the angles C or C is greater than 90° and 
the other is less than 90°, the vertex will be on CC produced 
through the vertex of the angle which exceeds 90°. (See 
Fig. 26, p. 366.) 

Note. If L" and L" represent the latitudes of C and C, respectively, 
the first formula of set [10] becomes 



GREAT CIRCLE SAILING 



371 



tan i(C + V) = C ° S \W ~ Jg cot * X rh 
sin \ (L + 2/ ) 

provided both places are on the same side of the equator. The student will 
find it a valuable exercise to express all of the preceding formulas of this 
section similarly. When both C and C" are south of the equator the 
polar distance must be reckoned from the south pole. 

To plot the Great Circle Track ox Mercator's Chart 

(1) By means of great circle charts. On these charts the 
great circle is represented, by a straight line ; hence, the lati- 
tude of each intersection with the meridians is known. The 
corresponding points may be marked on the meridians on 
Mercator's chart and a line traced through these points. 
This curved line will represent the great circle track. 

(2) By computing the latitudes corresponding to certain lon- 
gitudes (or vice versa) of the great circle, marking these points 
on Mercator's chart, and tracing the curve through them. 





fig. 32 



Fig. 33 



Example 1. Bequired the initial courses, the distance, and 
the latitude and longitude of the vertex of the great circle 
track joining C and C. Given : 

Lat. of C = 54°18'K Long, of C = 142° 38' E. 

Lat. of C = 7° 12' S. Long, of C" = 90° 0' W. 



372 



NAVIGATION 



Referring to the triangle CPC (Fig. 33), 

c =90°+ 7° 12' = 97° 12', 
c' = 90° - 54° 18' = 35° 42', 
X rf = P = (180° - 142° 38') + 90 c 



To find the initial courses : 
By [10], tan£(C + C") 



cos ^(c 



cos i (c + c') 



127° 22'. 



77C0tiA 



and 

%(c-c')= 30° 45' 

£(c+c') = 66° 27' 

*\* = 63°41' 



tan £(C 



sin 4- (c — c') 

C) = -r-3A 2 cot i x,;. 

y sm£(c + c') 2 d 



log cos£ (c - c') = 9.93420 

colog cos £(c+c')= 0.39843 

logcot£X d = 9.69425 



and 

To find the distance 



logtan£(C + C") =10.02688 

|(C + C") = 46° 46' 

.-. C = 62° 11', or N. 62° 11' E. 

C" = 31° 21', or N. 31° 21' W 



log sin \{c-c') = 9.70867 

colog sin £(c+c') = 0.03777 

logcot£X (/ = 9.69425 

log tan £ (C - C") = 9.44069 

£(C-C") = 15°25' 



-r, r- -.-.-> , ^ cos 4- (c H- c') . 

By [11], 0OS ^ = _J_L_^_ sm 



vKr 



l(c + c')= 66° 27' 

$(C + C') = 46° 46' 

$\ d = 63° 41' 



To find the latitude of the vertex : 

By [12], sin PV= sin c' sin C. 

c' = 35° 42' log sin c' = 9. 76607 
C = 62° 11' log sin C = 9.94667 
log sin PV = 9.71274 
PV = 31° 4' 
X of V = 90° - 31° 4' = 58° 56' N. 



logcosi(c + c') =9.60157 

colog cos | (C + C") = 0. 16433 

log sin $ \ d = 9.95248 

logcos£D = 9.71838 

£D = 58°29' 

.-. D = 116° 58' 

= 7018 miles. 

To find the longitude of the vertex 

% [13], 

cot CPV= cose' tan C. 

logcosc'= 9.90960 
log tan C = 10.27769 
log cot CPV= 10.18729 
CPV= 33° 1' 
X of C = 142° 38' E. 
X of V = 175° 39' E. 



GREAT CIRCLE SAILING 



373 



Example 2. The Cunard steamers in sailing from ISTew 
York to Liverpool keep approximately on the great circle 
between a point off the American coast in latitude 42° N., 
longitude 50° W., and Fastnet Rock off the Irish coast, in 
latitude 51° 24' K, longitude 9° 36' W. Required the initial 
course, the distance between these points, and the latitude 
and longitude of the vertex. 

In the triangle CPC (Fig. 34) let C represent 
the initial point, C" Fastnet, and P the pole, 
c' = 90° - 42° 0' 

= 48° 0' == colatitude of C, 
c = 90° - 51° 24' 
= 38° 36' = colatitude of C", 
\ d = P = 50° - 9° 36' = 40° 24' 

= difference of longitude of C and C". 




To find the initial course : 
By [10], tan i(C' + C) 



Fig. 34 



cos \ (c' — c) 
cos £ (c' + c) 



cot£A d , 



and 



tan£(C" 



sinHc--c) co 
' sin i(c'-f-c) 



$(c'-c)= 4° 42' 

i (c'+c) = 43° 18' 

£X rf = 20°12' 



log cos \(c'-c)= 9.99854 

colog cos -£(c'+c)= 0.13800 

log cot 1X^= 10.43424 



log tan i(C"+C) = 10.57078 
$(C'+C) = 74° 58' 
.-. C = 56° 59', or N. 56° 59' E. 



log sin ±(c'-c)= 8.91349 

colog sin £(c' + c) = 0.16379 

log cot £X d = 10.43424 

log tan $(C'-C)= 9.51152 

£(C"-C) = 17° 59' 



To find the distance : 

By [11], cos ±2) 

l(cf + c) = 43° 18' 

i(C+ C) = 74°58' 

iX d = 20°12 / 



cos \ (c' + c) 



sin i \ d . 



cos i(C' + C) 

log cos £(c' + c) = 9. 86200 

colog cosi (C + C) = 0.58606 

log sin i X rf = 9.53819 

log cos iD = 9.98625 

£D=14°21' 

.-. D = 28° 42' 

= 1722 miles. 



374 



NAVIGATION" 



To find the latitude of the vertex : 

By [12], sin P V = sine' sin C. 

log sine 7 = 9.87107 

log sin C = 9.92351 

logsinPF = 9.79458 



LofV 



PV = 38° 33' 

90° _ 38° 33' = 51° 27' N. 



To find the longitude of the vertex : 

By [13], cot CPV = cose' tan C. 

logcosc'= 9.82551 
log tan C= 10.18721 
log cot CPV= 10.01272 
CPV = 44° 10' 
X of C = 50° O'W . 
X of V = 5° 50' W. 



Note. £(C" + C) = 74° 57' 41" ; using this value, D = 1730 miles. 
From Sandy Hook to the company's position is 1088 miles; thence 
to Fastnet by great circle, 1730 miles; total, 2818 miles. The routes 
recommended by Captain Maury are mainly great circle routes. 



EXERCISE Vm 

1. Find, the elements (initial courses, distance, and latitude 
and longitude of the vertex) of the great circle track between 
the Lizard, in latitude 49° 58' K, longitude 5° 12' W., and the 
Bermuda Islands, in latitude 32° 18' N., longitude 64° 50' W. 

2. Find the elements of the great circle track between 
Boston (Minot's Ledge lighthouse), in latitude 42° 16' K, 
longitude 70° 46' W., and Cape Clear, in latitude 51° 26' N., 
longitude 9° 29' W. (Take i \ d = 30° 39'.) 

3. Find the elements of the great circle track between 
Vancouver Island, in latitude 50° N., longitude 128° W., and 
Honolulu, in latitude 21° 18' K, longitude 157° 52' W. 

4. Find the elements of the great circle track between Cape 
Clear, in latitude 51° 26' N ., longitude 9° 29' W., and Sandy 
Hook, in latitude 40° N., longitude 74° W. 

5. Find the elements of the great circle track between the 
Lizard, in latitude 49° 58' K, longitude 5° 12' W., and Cape 
Frio, in latitude 23° S., longitude 42° W. 



GREAT CIRCLE SAILING 375 

6. Find the elements of the great circle track between Cape 
Frio and Cape of Good Hope, in latitude 34° 20' S., longitude 
18° 30' E. (Reckon from the nearest pole.) 

7. Find the elements of the great circle track between 
Grand Port, Mauritius, in latitude 20° 24' S., longitude 57° 
47' E., and Perth, in latitude 32° 3' S., longitude 115° 45' E. 

8. Find the elements of the great circle track between A, 
in latitude 16° 38' N., longitude 70° 55' W., and B, in latitude 
48° 2' K, longitude 4° 35' W. 

9. A ship sails from A, in latitude 40° S., longitude 148° 30' 
E., to B, in latitude 12° 4' S., longitude 77° 14' W. Compare 
the great circle track and the rhumb line between A and B. 



CHAPTER III 
NAUTICAL ASTRONOMY 

SECTION XIX 

the observed altitude of a heavenly body 

The Sextant 

The sextant is an instrument used by navigators for deter- 
mining the angular distance between two heavenly bodies, or 
the angular altitude of a heavenly body above the horizon. 

It is held in the hand, and by a turn of the wrist the 
observer can keep the plane of the instrument in the plane of 
the two bodies whose angular distance he wishes to measure. 
The angular distance is measured not by pointing first on one 
and then on the other, but by sighting them both simulta- 
neously and in apparent coincidence. This can be done even 
when the observer has no fixed position or stable footing. 

This instrument is shown in outline in Fig. 35. It consists 
of a frame ABC in the form of a sector of radius 5 to 8 inches 
whose arc AB is 60° in extent, but divided into 120 equal 
parts reckoned as degrees, and each so-called degree is divided 
into six equal parts. CE is a movable radius which carries 
at C a mirror called the index glass, whose plane is perpen- 
dicular to the plane of the sextant ; and at E an index which 
moves over the arc AB. At D is fixed the horizon glass, also 
perpendicular to the plane of the sextant and parallel to CA ; 
that is, parallel to the index glass when the index is at (A). 

376 



OBSERVED ALTITUDE OF A HEAVENLY BODY 377 

Only one half of the horizon glass is silvered, the upper half 
being left transparent. F is a telescope fixed to ('A, with its 
axis directed nearly to the line which separates the silvered 
and unsilvered parts of the horizon glass. 



s*. 




>- 



Fig. 35 



To determine the angular distance between two heavenly 
bodies S and S' (Fig. 35), the sextant is held in the hand so 
that a ray of light from S' passes through the unsilvered part 
of the horizon glass, enters the telescope, and is seen as if by 
direct vision. The arm CE is then moved so that a ray of 
light from S being reflected by the mirrors at C and D, enters 
the telescope, taking the path SCDF. The movement of CE 
is continued until the image of S appears to be in contact with 
S', when the arc AE gives the angular distance required. 

The angle between the objects whose images coincide is twice 
the angle between the mirrors. 



378 NAVIGATION 

By the law of reflection, a = a, and b = b (Fig. 36). 
Q .= a - b. 
M = 2 a - 26 = 2 Q = 2 Q' = 2 ACE. 
Half degrees on AB are marked as degrees. Hence, AE is the required 
reading. 

If S' is the point on the sea horizon directly under S, the 
arc AE will give the observed altitude of S. The altitude of 

the sun is measured by point- 

N^ ing the telescope at the sea 

a y jfi — — ;>-""' horizon, which appears like a 

P-"'" a Afk K ^ l me i n ^ ne ^ e ^ °^ the telescope, 

# > _._.\/::M \ --.:. M and by moving the index arm till 

s / \ \ \ one e( ^S e °f the reflected image 

/ \ \ \ of the sun coincides with the 

/ \ \ \ horizon. The arc read from the 

/ \ , \ sextant at this time is the sun's 

\> — ~r r ^^ altitude. When the altitude 

_, oe of the sun is observed it has 

to be corrected for index error, 
refraction, dip, parallax, and semidiameter. 

The Quintant has an arc of 72°, and is therefore capable of 
measuring angles up to 144°. In other respects it is like the 
sextant. 

The Octant (commonly, but improperly, called the quadrant) 
has an arc of 45°, and is therefore capable of measuring angles 
up to 90°. In general construction it is similar to the sextant, 
for which it is a cheap substitute. It is much more easily 
handled, and is yet sufficiently accurate for many purposes. 
In Fig. 35 the arc AB, extending from zero to the left, is 
called the arc proper ; the arc to the right of zero is called the 
arc of excess. The index is on the arc or off the arc according 
as it is on the arc proper or on the arc of excess. 

When the index glass is parallel to the horizon glass the 
index should be at 0°. If it is not, the arc between the zero 



CORRECTIONS OF THE OBSERVED ALTITUDE .'379 

point and the index is called the index error. The index 
correction is readily discovered as follows: 

Bring the direct and reflected images of a star into coin- 
cidence and read off the arc. The index correction is 
numerically equal to this reading, and is positive or negative 
according as the reading is on the right or left of the zero. 
The sea horizon may be used instead of a star. 

In a perfect instrument the vernier should read strictly 0° 
when the mirrors are parallel; but however nicely a sextant 
may be constructed, there is always a small amount of error 
in the parallelism of the index and horizon glasses. 



SECTION XX 

CORRECTIONS OF THE OBSERVED ALTITUDE 

Refraction 

The air which surrounds the earth gradually decreases in 
density as we ascend from the surface of the earth. At the 
height of four miles its density is only 
half as great as at the earth's surface. 
According to a fundamental law of 
optics, if a ray of light passes from a 
rarer to a denser medium, it is bent 
toward the perpendicular to the surface 
which separates the two media ; but if 
it passes from a denser to a lighter 
medium, it is bent from that perpen- 
dicular. This law is true when the 
surface is curved as well as when it 
is plane. 

Let S (Fig. 37) represent the true position of a star, O the 
position of the observer on the surface of the earth, and PP' 
the upper limit of the atmosphere. 




Fig. 37 



380 NAVIGATION 

A ray of light from S meeting the atmosphere at P will be 
bent toward the perpendicular to the arc PP' at P. As the 
density of the atmosphere increases from P to O, PO will be 
a curve, and the star will be seen in the direction OS' tangent 
to PO at 0, the last direction of the ray being that of a tangent 
to the curved path at the eye of the observer. These laws are 
here assumed. The facts and reasoning on which they depend 
belong to works on Optics. 

Astronomical refraction is the difference of direction of the 
two lines PS and OS' (Fig. 37). 

The effect of refraction is to increase the apparent altitude 
of a celestial object, and is greatest when the body is near the 
horizon, and decreases as the altitude of the body increases. 

At the zenith the refraction is nothing. The less the alti- 
tude, the more obliquely the rays enter the atmosphere and 
the greater is the refraction. 

Table XIII, constructed partly from observation and partly 
from theory, gives the correction for mean refraction corre- 
sponding to every apparent altitude from 0° to 90°. This 
correction must evidently be subtracted from the apparent 
altitude. 

The effect, therefore, of refraction is to increase the altitude 
of a star without affecting its azimuth. 

The refraction varies slightly according to the condition of 
the atmosphere, but this variation is not taken into account in 
this chapter. 

Dip of the Horizon 

Let BH" (Fig. 38) represent the level of the sea, the posi- 
tion of the observer at the height BO above the sea level, C 
the centre of the earth, and H" the farthest point on the sur- 
face of the earth visible from O. On account of refraction 
the visual ray H"0 will be a curve concave toward C, and H" 
will be seen in the direction of OH' tangent to H"0 at 0, and 




CORRECTIONS OF THE OBSERVED ALTITUDE 381 

OH' will represent the visible horizon. Let OH, perpendicular 
to OC, represent the true horizon, from which the apparent 
altitude of the body is 
estimated. 

When we measure the 
altitude of S, the sun, we 
measure the angle SOH'; 
but the altitude we re- 
quire is the angle SOH: 
hence, we measure too 
much by the angle 
HOW. The visible hori- 
zon has dipped below 

the rational horizon by the angle II OH', called the Dip of the 
Horizon. We see then that the dip must be subtracted from 
the observed altitude, and that its amount depends entirely 
upon the height of the observer. 

The dip in minutes of arc is very nearly proportional to 
the square root of the height of the observer in feet above 
the sea level. 

This element is found in Table XV, which gives the dip 
corresponding to heights up to 100 feet. The angle HOH' is 
small even up to 100 feet, being for that height 9' 48". Navi- 
gators on the decks of their vessels are rarely over 25 feet 
above the sea level. 



Fig. 38 



Parallax 

Observations of the celestial bodies comparatively near the 
earth, when made at different places, do not give the same 
result, because they are viewed from different points. To 
correct this difference, astronomers have agreed to reduce the 
observations to what they would be if they had been made at 
the centre of the earth. 



382 



NAVIGATION 



The difference between the positions of a heavenly body as 
seen from the centre of the earth and a point on its surface at 

the same time is called 
■ // geocentric parallax. 

Parallax may also be 
regarded as the angle at 
the celestial body made 
by two lines drawn from 
it, one to the observer and 
the other to the centre 
of the earth. Parallax 
therefore depends upon 
the distance of the celes- 
tial body ; the nearer the 
celestial body the greater 
the parallax. The moon 
has the greatest parallax, 
then the nearest planets, and then the sun. 

Let S (Fig. 39) represent the celestial body, the position 
of the observer, C the centre of the earth, and Z the zenith of 
the observer. 

The geocentric parallax is the angle 

s = zos - zcs. 

ZOS is the apparent zenith distance, and ZCS the geo- 
centric or true zenith distance. It is evident that geocentric 
parallax decreases as the altitude of the body increases. 

When a body is between the horizon and zenith the parallax 
is called parallax in altitude. 




Fig. 39 



Let 



P == the parallax in altitude, 

Z = ZOS, the apparent zenith distance, 

R= CO, 

D = CS. 



CORRECTIONS OF THE OBSERVED ALTITUDE 383 

In the triangle COS, 

sin S sin COS sin P sin Z 

= , or = - — — • 

CO CS R I) 

Hence, sin P = - — - — - • (a) 

Since P is always a small angle, the arc may be taken for 
its sine, and the circular measure of the arc for the arc itself. 

P R sin Z 

^erefore, —^ = ——, 

and P" = 206265" II *™ Z • 

That is, parallax varies directly as the sine of the zenith 
distance, and inversely as the linear distance (in miles) of 
the body. 

If the body is on the horizon as at //, the angle H is called 
the horizontal parallax = P'. In this case the angle Z = 90°, 
and the preceding formula becomes 

sinP' = -■ 

Substituting sin P' for — in formula (a), and putting h for 
SOH, whence Z = 90° - h, 
we have sin P = sin P' cos h. 

Since P and P' are very small, their sines are nearly pro- 
portional to the angles. 

Hence, P = P' cos A. 

That is, the parallax in altitude is equal to the horizontal 
parallax multiplied by the cosine of the apparent altitude. 

It is from this expression that the table for parallax in 
altitude is constructed. 

The sun's mean horizontal parallax is about 8.85", whence 
the sun's parallax in altitude may be computed. 



384 



NAVIGATION 



This element is given in Table XVII, calculated for all 
altitudes of the sun from 0° to 90°. There must, of course, be 
separate tables for sun, moon, and planets. 

As parallax causes the body to appear lower than if it 
were seen from the centre of the earth, the correction for 
parallax must evidently be added to the apparent altitude. 

The parallax of the fixed stars is practically zero. The 
horizontal parallax for the planets varies, because they are 
at different distances from the earth at different days dur- 
ing the year. In the same way the horizontal parallax of 
the moon varies. These elements are given in the Nautical 
Almanac for intervals of five days. As the equatorial semi- 
diameter of the earth is larger than any other semidiameter, 
so also is the equatorial horizontal parallax larger than any 
other horizontal parallax. 

The Sun's Apparent Semidiameter 



Let C (Fig. 40) represent the centre of the earth, C the 
centre of the sun, and CP a tangent to the surface of the sun. 
Angle PCC is the semidiameter of the sun as viewed 
from C. 

Let CC = D, OC = R, PC' = r, 
angle PCC = S = the sun's semidi- 
ameter. 

In the right triangle CPC, 
r 




sin S = 



T) 



Fig. 40 



The formula for horizontal parallax 

, R ^~ R 

is sin P = — > or D = — — — ; 
D sin P' ' 



whence, 

since S and P' are very small 



T V 

sin S = — sin P\ or S = — P', nearly, 
R R 



CORRECTIONS OF THE OBSERVED ALTITUDE 385 

Hence, the semidiameter of the sun may be found from its 
horizontal parallax. 

Note. The semidiameter of a heavenly body when on the horizon, 
as viewed from the centre of the earth, is called the horizontal semi- 
diameter. The semidiameter is sensibly the same whether viewed from 
C or 0, provided the body is on the horizon, since in this position CC and 
OC are nearly equal. As the altitude of the body increases, OC decreases, 
and the semidiameter as viewed from O increases. This diurnal augmen- 
tation of the semidiameter is sensible in the case of the moon, but not 
in the case of the sun. In the case of the moon there is also a monthly 
augmentation owing to the varying distance of the moon from the earth. 

The sun's semidiameter is given for every day in the year 
in Table XII. It is greatest when the sun is nearest the 
earth, about Jan. 1, and least when the sun is farthest from 
the earth, about July 1. The Nautical Almanac contains the 
semidiameters as well as the horizontal parallaxes of the sun. 
moon, and planets. 

In order to find the true altitude of the sun's centre, the 
semidiameter is added to the apparent altitude of the lower 
limb and subtracted from the apparent altitude of the upper 
limb. 

Applying the Corrections 

The altitude of the limb is read from the sextant. 

The index error being applied, the result is the observed 
altitude above the sea horizon. 

The correction for dip being applied, the result is the 
apparent altitude of the limb, referred to the true horizon. 

The true altitude of the body is the altitude of its centre 
as viewed from the centre of the earth, and is obtained by 
applying the corrections for refraction, parallax, and semi- 
diameter to the apparent altitude of the limb. 

The fixed stars have neither parallax nor semidiameter. 



386 NAVIGATION 

The following symbols will be used in this chapter : 

O the sun, 

© the sun's centre, 

the sun's upper limb, 

Q the sun's lower limb. 

Example 1. The observed altitude of a star was 18° 20 f 
30"; index error, + 1' 17"; height of eye, 18 feet. Find the 
true altitude. 

Solution I 

Reading of sextant =18° 20' 30" 

Index error = +1' 17" 

Observed altitude 

Dip (Table XV) 

Apparent altitude 

Refraction (Table XIII) 

True altitude = 18° 14' 44" 

Solution II 

Reading of sextant = 18° 20' 30" f Index error = + 1' 17" 
Correction = — 5' 46" ^ Dip = — 4' 9" 

^Refraction — — 2 r 54" 

True altitude = 18° 14' 44" Correction = - 5' 46" 

The corrections are properly made in a certain order in the first solu- 
tion ; in the second solution the algebraic sum of the corrections is applied 
to the observed altitude. It is evident that the correction for refraction 
might differ by the two methods, since in one case it is based on an 
altitude of 18° 17' 38", and in the other on an altitude of 18° 20' 30"; 
but in any case the difference would be so slight as to be practically 
unimportant in sea observations. The second method is commonly used 
in practice. 

Example 2. Aug. 9, 1895, the mean of observed altitudes 
of the sun's lower limb was 26° 0' 2"; index error, — 0' 17"; 
height of eye, 18 feet. Find the sun's true altitude. 





18° 


21' 


47" 






-4' 


9" 


= 


18° 


17' 


38" 






-2' 


54" 



CORRECTIONS OF THE OBSERVED ALTITUDE 387 



Solution I 

Reading of sextant 

Index error 

Observed altitude 

Dip 

Apparent altitude O 

Refraction - 1' 59'H 

Parallax + 0' 8"| 

True altitude © 

Semidiameter (Table XII) 

True altitude © = 26° 9' 34' 

Solution- II 

Reading of sextant = 26° 0' 2" 



Correction = 0' 32' 



= 26° 


ry 


2" 


= 


- 0' 


17" 


= 25° 


o<r 


45" 


— 


-4' 


9" 


= 25° 


55' 


36" 


= " 


- V 


51" 


= 25° 


63' 


45" 


= + lo- 


49" 



Semidiameter 


= + 15' 49" 


Parallax 


= + <y 8- 


Index error 


= - C 17" 


Dip 


= - 4' 9" 


Refraction 


= - r 59" 


Correction 


= + 9' 32" 



True altitude = 26° 39' 4' 

Do not interpolate for refraction, but take the nearest number from 
the table. 

EXERCISE IX 

From the following data compute the true altitude : 

1. Obs. alt. star 25° 6' 10", index cor. + 1' 15", eye 17 feet. 

2. Obs. alt. star 15° 20' 25", index cor. - 2' 20", eye 16 feet. 

3. Obs. alt. 18° 17' 30", index cor. + 0' 18", eye 18 feet. 

4. Obs. alt. © 30° 12' 40", index cor. 0' 0", eye 19 feet. 

5. Obs. alt. O 56° 25' 20", index cor. - 1' 20", eye 17 feet. 

6. Obs. alt. © 60° 10' 10", index cor. - 2' 15", eye 18 feet. 

7. Obs. alt. © 31° 24' 35", index cor. 0' 0", eye 18 feet. 

8. Obs. alt. 26° 17' 20", index cor. + 2' 15", eye 18 feet. 

9. Obs. alt. 20° 35' 30", index cor. + 0' 18", eye 16 feet. 
10. Obs. alt. © 36° 12' 10". index cor. + 0' 25", eye 20 feet. 



388 NAVIGATION 

SECTION XXI 
TIME 

Apparent Solar Time is measured by the daily motion of 
the sun. 

An Apparent Solar Day is the interval of time between two 
successive transits of the sun over the same meridian. This 
interval is not always the same, for two reasons : 

(1) The sun moves in the ecliptic, which is oblique to the 
equator. 

(2) The rate of the sun's motion in the ecliptic varies, being 
greatest when the earth is at perihelion, and least when the 
earth is at aphelion. 

Hence, a clock or a chronometer cannot be regulated to 
follow the true sun. 

The Mean Sun is an imaginary sun supposed to move in the 
equator at a uniform rate equal to the mean rate of the true 
sun. 

Mean Solar Time is the hour angle of the mean sun, because 
it is the number of hours from the meridian or noon. 

Apparent Solar Time is the hour angle of the true sun. 

The Equation of Time is the difference between mean time 
and apparent time. It never exceeds about 16 minutes. 

Sidereal Time is measured by the apparent daily motion of 
the fixed stars. The sidereal time at any moment may be 
defined as the hour angle of the vernal equinox. 

The Sidereal Day is the interval between two successive tran- 
sits of the vernal equinox over the same meridian. It is about 
3 minutes 56 seconds shorter than the mean solar day, and is 
divided id to 24 hours numbered from to 24. 

The Civil Day is the interval of time between two successive 
midnights. It is divided into two periods of 12 hours each, 
the first of which is marked a.m. and the second p.m. 



LONGITUDE AXD TIME 



389 



The Astronomical Day begins at noon of the civil day of the 
same date; that is, 12 hours after the commencement of the 
civil day. It is divided into 24 hours numbered from to 24. 

To convert civil time into astronomical time : 
If the given time is a.m., deduct 1 day, add 12 hours, and 
omit the a.m. ; if the given time is p.m., simply omit the p.m. 

To convert astronomical time into civil time : 

If the hours of the given time are less than 12, simply affix 
p.m. ; if the given hours are 12 or more, add 1 day, subtract 
12 hours, and affix a.m. 

EXERCISE X 



1 


Astronomical Time 


Civil Time 


d. li. m. 
July 8 7 


s. 
10 


Required 


2 


Mar. 7 12 25 


30 


Required 


3| Jan. 1 18 10 


10 


Required 


4 


Dec. 31 15 





Required 


5 


Feb. 2 8 4 


30 


Required 
d. h. m. s. 


6 


Required 




July 111 8 25 a.m. 


7 


Required 




Mar. 2 11 56 56 p.m. 


8 


Required 




Aug. 31 10 8 20 p.m. 


9 


Required 




Sept. 1 12 15 a.m. 


10 


Required 




Jan. 1 10 41 56 a.m. 



SECTION XXII 

LONGITUDE AND TIME 

Since the earth revolves on its axis once in 24 hours, the 
sun appears to move over the 360° of the equator in the same 
time. Hence, the longitude of any place is proportional to 



390 NAVIGATION 

the time required by the sun to move over an arc of the 
equator included between the meridian from which longitude 
is reckoned and the meridian of the place; and longitude may 
be expressed either in tim,e or arc. 

Since 360° = 24 hours, 

15° = 1 hour, "| f 1° = 4 minutes, 

15' = 1 minute, > or ^ 1' — 4 seconds, 
15" = 1 second, J ll" = T ^ second. 

Either of the above tables enables us to convert longitude 
in time to longitude in arc, and vice versa. 

SECTION XXIII 

THE GREENWICH DATE 

The Greenwich date is the time at Greenwich correspond- 
ing to the local time elsewhere. Greenwich time should 
always be expressed astronomically. 

To find the Greenwich date, having given the ship time and the 
longitude of the ship : 

(1) Express the ship time astronomically (Sect. XXI, 
p. 389). 

(2) Convert longitude in arc into longitude in time 
(Sect. XXII, p. 390). 

(3) If the ship is in west longitude, add the longitude in 
time ; if the ship is in east longitude, subtract the longitude 
in time. 

The resulting Greenwich date is expressed in apparent time 
or mean time according as the local time is apparent or mean. 

It is evident that local time can be found from Greenwich 
time by reversing these steps. 

Example 1. June 4, at 6 hours 10 minutes p.m., apparent 
time, at ship in longitude 30° 15' E. Find the Greenwich date. 



THE GREENWICH DATE 



391 



d. h. m. 

Ship date (expressed astronomically) = June 4 6 10 

Longitude in time = — 2 1 

Greenwich date (apparent time) = June 4 4 9 

15 )30° 15' = longitude in arc. 
2 h. 1 m. = longitude in time. 



Example 2. Aug. 21, at 8 hours 30 minutes 10 seconds a.m.. 
mean time, at ship in longitude 90° 30' W. Find the Green- 
wich date. 

d. h. m. s. 

Ship date (expressed astronomically) = Aug. 20 20 30 10 



Longitude in time 
Greenwich date (mean time) 



+ 6 2 



Aug. 21 2 32 10 



Example 3. March 20, the sun on the meridian of the place 
in longitude 160° 15' W. Find the Greenwich date. 



d. h. m. s. 

Ship date (expressed astronomically) = Mar. 20 

Longitude in time = +10 41 

Greenwich date (apparent time) = Mar. 20 10 41 



EXERCISE XI 





Local Civil D 


VIE 


Longitude 


Greenwich Date 


d. h. m. 


s. 






1 May 4 6 12 


15 P.M. 


170° 50' 0"W. 


Required 


2 


July 31 11 12 


30 A.M. 


40° 20' 0"W. 


Required 


3 


Aug. 1 2 10 


15 A.M. 


80° 40' 45" W. 


Required 


4 


Mar. 2 10 20 


P.M. 


30° 45' 0"E. 


Required 


5 


Mar. 25 11 • 8 


P.M. 


100° 25' 30" AY. 


Required 
d. h. m. s. 


6 


Required 




25° T 0"W. 


Dec. 30 19 47 28 


7 


Required 




179° 0' 0"W. 


July 4 23 51 


8 


Required 




179° 0' 0" E. 


July 3 23 59 


9 


Required 




45' 0"E. 


May 19 19 40 20 

1903 


10 


Required 




2° 10' 0"E. 


Dec. 31 15 8 



392 NAVIGATION 

SECTION XXIV 
THE CHRONOMETER 

A chronometer is a large watch of peculiar construction, 
made especially for the use of navigators. It is suspended 
by gimbals, as a compass bowl is, and the object is to keep 
the face of the chronometer horizontal. It is kept in a box 
in the navigator's cabin and preserved as much as possible 
from changes of temperature and jars, for upon it depends 
the means of finding the longitude. 

Unless otherwise specified, chronometers are supposed to be 
set to Greenwich mean time. 

The error of chronometer on mean time at any place is the 
difference between the mean time at that place and the time 
indicated by the chronometer. 

The Error of Chronometer on mean time at Greenwich is the 
difference between mean time at Greenwich and the time by 
chronometer. The error is fast or slow according as the chro- 
nometer is in advance of or behind the mean time at Greenwich. 

The Rate of Chronometer is the daily change in its error; 
it is gaining or losing according as the chronometer is going 
too fast or too slow. 

Every chronometer has an error which is accurately deter- 
mined, generally at an observatory, where a memorandum is 
kept of its career as a timekeeper. The purchaser always 
receives a certificate stating how much the chronometer was 
too fast or too slow for Greenwich mean time at a specified 
date; also how much it gains or loses on the average in a 
day, that is, its daily rate. As there can be no guarantee that 
the rate will not change, it is important to ascertain the rate 
at subsequent intervals. 

To determine the error of chronometer on Greenwich mean 
time the longitude of the place must be known, whence the 



THE CHRONOMETER 393 

Greenwich mean time may be found (Sect. XXII I, p. 390); 
this compared with the chronometer set to Greenwich time 
gives the error. 

The rate may be found by determining the error on different 
days and noting the daily change, supposing it to be uniform. 

Example 1. Feb. 12 the error of chronometer was 9 minutes 
19.6 seconds slow, and Feb. 25 the error was 10 minutes 30.2 
seconds slow. Find the rate. 



Feb. 12, slow 


= 9 19.6 


Feb. 25, slow 


= 10 30.2 


Change for 13 days 


= 1 10.6 




60 




13)70.6 



Change for 1 day, or rate = 5.4 s. losing. 
Sinoe the chronometer was slow, and the error increasing, the rate 
was losing. 

Example 2. April 8, p.m., an observation was made, when 
the time was April 8, 4 hours 10 minutes 22 seconds by 
chronometer, which was 18 minutes 10 seconds fast on Green- 
wich mean time on Jan. 1, and 16 minutes 17 seconds fast on 
Jan. 31. Find the Greenwich date by chronometer. 

m. ». 

Jan. 1, fast = 18 10 

Jan. 31, fast = 16 17 

Change for 30 days = 1 53 
60 
30)113 
Rate = 3.8 s. losing. 

Since the chronometer was fast, and the error decreasing, the rate 
was losing. d h m g 

Time by chronometer = April 8 4 10 22 
Original error = — 16 17 

8 3 54 5 
Accumulated rate + 4 15.3 

Greenwich date = April 8 3 58 20.3 

255.25 s. = 4 m. 15.3 s. 255.25 



d. h. 


s. 


Feb. 28 


3.8 


March 31 


228.0 60 


April 8 4 
67 4 


26.6 7 
0.38 0.1 


67.17 


0.27 0.07 



228.0 


= 3.8 


X 


60 


26.6 


= 3.8 


X 


7' 


0.38 


= 3.8 


X 


0.1 


0.27 


= 3.8 


X 


0.07 



394 NAVIGATION 

The error on Jan. 31 is 16 m. 17 s. fast, which we deduct. From 
Jan. 31 to April 8d. 3h. 58 m. is nearly 67 d. 4h. 
= 67.17 d. The error for one day is 3.8 s.; 3.8 s. 

hence, for 67.17 d. the error is 67.17 x3.8s. 
= 255.3 s. = 4 m. 15.3 s. As the chronometer is 
losing time, this accumulated rate (4 m. 15.3 s.) 
must be added, and the result is the required 
Greenwich date. 255.25 = 3.8 x 67.17 

The method of multiplying 3.8 s. by 67.17 
employed on page 393 will be made plain by the arrangement of the 
work shown in the margin. 

SECTION XXY 
THE NAUTICAL ALMANAC 

The American and British Nautical Almanacs give, at 
equidistant instants of Greenwich time, the apparent right 
ascensions and declinations of the sun, moon, planets, and 
certain fixed stars ; the semidiameters of the sun and moon ; 
the equation of time ; and so on. 

Table XII contains the data from the Nautical Almanac 
which are necessary for the solution of the exercises in this 
chapter. 

To reduce the sun's declination by means of the hourly 
difference : 

Table XII gives the sun's declination at apparent and mean 
noon, Greenwich time, and the hourly change in declination 
for every day in the year. 

If the declination at noon is required, it may be taken at 
once from the table. The declination at any other time may 
be found by simple interpolation ; that is, by assuming that 
the declination varies as the time. The column headed 
"Diff. for 1 hour" contains the average change for one hour. 
This multiplied by the number of hours gives the correction 
required. 



THE NAUTICAL ALMANAC 395 

Note. In Table XII the difference of declination for 1 hour is 
marked + or — according as the sun is moving northward or south- 
ward, but in the following examples '-Diff. for 1 hour" is marked 
+ or — according as the element to which it is to be applied is increasing 
or decreasing numerically. 

Example 1. Find the sun's declination for 1895, Jan. 
7 days 10 hours, apparent time, Greenwich date. 

O's dec. Jan. 7 d. h. = 22° 22' 25.8" S. Diff. for 1 h. = - 19.23" 
-a 7 12.3" xlO 

O's dec. Jan. 7 d. 10 h. = 22° 19' 13.5" S. Diff. for 10 h. =- 192.3" 

= -3' 12.3" 
The table gives the declination for Jan. 7 d. h. = 22° 22' 2-3.8" S., and 
the hourly difference = 19.23", which is marked — because the declina- 
tion is decreasing. For 10 h. the declination is 10 x 19.23" = 192.3" 
= 3' 12.3", which must be subtracted from the declination at noon. 

Example 2. Find the sun's declination for 1895, Feb. 
12 days 17 hours 12 minutes 10 seconds, mean time, Green- 
wich date. 

O's dec. Feb. 13 d. h. = 13° 20' 27.0" S. Diff. for 1 h. = - 50.49" 

+ 5' 43.3" x 0.8 

O's dec. Feb. 12 d. 17 h. Diff. for 6.8 h. = 343.33-'' 

12 m. 10 s. = 13° 26' 10.3" S. . = 5' 43.3" 

In this example the given date is nearer Feb. 13 d. h. than Feb. 12 d. 
Oh.; hence, we take from the table the declination for the former date, 
Feb. 13d. Oh. 

Feb. 12 d. 17 h. 12 m. 10 s. = 6 h. 47 m. 50 s. (= 6.8 h.) before noon 
Feb. 13. The difference is found to be 5' 43.3". which must be added to 
the declination at noon. 

Example 3. 1895, May 5, apparent noon, longitude 57° W. 
Eequired the sun's declination. h 

Local time = May 5 

Longitude in time = 3 48 

Greenwich date = May 5 3 48 

O's dec. May 5 d. h. = 16° 15' 42.5" Diff. for 1 h. =+ 42.76" 

+ 2' 42.5" x 3.8 

O's dec. May 5 d. 3 h. 48 m. = 16° 18' 25" Diff. for 3.8 h. = 102.488" 

= + 2' 42.5" 



396 NAVIGATION 



TO REDUCE THE EQUATION OF TlME 

Example 4. Greenwich date,' mean time, 1895, March 
2 days 3 hours 10 minutes. Find the equation of time. 



Eq. of time March 2 d. h. =-12 19.64 


Diff. forlh. =-0.514 


- 1.63 


xS.lf 


Eq. of time March 2 d. 3h. 10 m. = - 12 18.01 


Diff. for 3. If h. =-1.6277 



Table XII gives the equation of time corresponding to mean time, 
March 2 d. h. = 12 m. 19.64 s. The — sign indicates that the equation 
of time is to be subtracted from mean time to obtain apparent time. 

The difference for 1 hour is also found in the table and marked — , 
because the equation of time is decreasing. The difference for 3 h. 10 m. 
is found to be 1.63 s., which is subtracted from the equation of time at 
noon. 

This method is not strictly correct, because the change in declination 
is not nniform ; but it will be found accurate enough in most cases, and 
will be used in this chapter. 

EXERCISE XH 

Find the sun's declination and the equation of time corre- 
sponding to the following Greenwich dates : 

d. h. m. s. 

1. 1895 Jan. 7 3 apparent time. 

2. 1895 Aug. 1 6 12 20 apparent time. 

3. 1895 May 5 10 25 apparent time. 

4. 1895 Aug. 7 15 12 apparent time. 

5. 1895 Dec. 4 6 18 apparent time. 

6. 1895 July 23 20 16 40 apparent time. 

7. 1895 Nov. 13 6 apparent time. 

8. 1895 Oct. 12 5 12 apparent time. 

9. 1895 June 7 3 18 mean time. 
10. 1895 Feb. 3 9 15 mean time. 



TO FIND THE HOUR ANGLE .397 

SECTION XXVI 

TO FIND THE HOUR ANGLE OF A HEAVENLY BODY 

Let t = the hour angle of the heavenly body. 
L = the latitude of tjie observer. 
d = the declination of the heavenly body. 
h = the true altitude of the heavenly body. 
p = 90° ± d 

= the polar distance of the heavenly body. 
S = i(L+p + h). 
R = i(L+p-h) 

=:ilL+p + h)-h 

= S - It. 

The general formula for the hour angle of a heavenly body 
in terms of its polar distance and altitude and the latitude of 
the observer is 

sin L t = ± [cos S sin R sec L esc p]\ 

(Spherical Trigonometry, LXX, p. 192.) 

Example 1. The true altitude of a heavenly body is 
14° 50' 42"; its declination, 16° 20' 5" S. ; and the latitude of 
the observer, 37° N. Find the hour angle. 

£= 37° 

p = 106° 20" 5" log cos S = 9.27708 

h = 14° 50 7 42" log sin B = 9. 95456 

2 8 = 158° 10' 47" log sec L = 0.09765 

S = 79° 5' 24" log esc p = 0.01789 

R= 64° 14' 42" 2 )19.34718 
log sin it = 9.67359 

^ = 28° 8' 22.5" 
.-. t = 56° 16' 45" 
= 3 h. 45 m. 7 s. 



308 



NAVIGATION 



SECTION XXVII 

TO FIND THE LOCAL TIME 

The local apparent time is the hour angle of the true sun, 
if the sun is west of the meridian; if the sun is east of the 
meridian, the local apparent time is the difference between 
24 hours and the hour angle. If, however, we count the hour 
angle around the circle, the sun's hour angle is the apparent 
time. 



Example 1. At sea, Sept. 12, a.m., in latitude 18° 12' S., 
Greenwich mean time, 1895, Sept. 11 days 14 hours 55 min- 
utes 50 seconds; observed altitude of CD, 26° 18' 20"; index 
error, + 31"; height of eye, 18 feet. Eequired the local mean 
time. 

Sept. lid. 14 h. 55m. 50s. = 9 h. 4m. 10s. (=9.07 h.) before noon, Sept. 12. 

To reduce the sun's declination and find the polar distance : 
O's dec. Sept. 12 d. h. = 
Change for 9.07 h. = 

O's dec. Sept. 11 d. 11 h. 

55 m. 50 s. = 



4° 10' 47.1" N. 


Diff. 


for 1 h. = - 57.32" 


+ 8' 39.9" 


Diff. 


for9h. = 515.88" 




Diff. 


for 0.07 h. = 4.01" 


4° 19' 27" N. 


Diff. 


for 9.07 h. = 519.89" 


90° 




= 8' 39.9" 


91° 19' 27" 







To find the true altitude : 

Reading of sextant = 26° 18' 20' 
Correction = +10' 29' 



True altitude © = 26° 28' 49' 



Semidiam. = + 15' 56' 
Index error = -f- 31' 
Parallax = + 8' 

Dip =- 4' 9' 

Refraction = — V 57' 
Correction = + 10' 29' 



To reduce the equation of time : 

m. s. 

Eq. of time Sept. 12 d. h. =3 44.76 
Change for 9.07 h. = - 7.93 

Eq. of time Sept. 11 d. 14 h. 

55 m. 50 s. = 3 36.8 



Diff. for 1 h. = + 0-874 

Diff. for 9 h. = 

Diff. for 0.07 h. = 

Diff. for 9.07 h. - 



7.866 
0.061 
7.927 



TO FIND THE LOCAL TIME 



399 



To find the hour angle of the sun : 

sin \ t = ± [cos S sin R sec L esc p]K 

log cos S = 9.54428 

log sin R= 9.83396 

log sec L = 0.02229 

log esc p = 0.00123 



L = 18° 12' 0" 
p - 94° 19' 27" 
A = 26° 28' 49" 



2S= 139° 0' 16" 
S = 69° 30' 8" 
E = 43° 1' 19" 



2 )19.40176 
log Bin it- 9.70088 

£$=30° 8' 46" 

.-. t = 60° 17' 32" 
= 4 h. 1 m. 10 s. 



Local apparent astronomical time = Sept. 11, 
Equation of time = 

Local mean astronomical time = Sept. 11, 
Local mean civil time = Sept. 12, 



24 





4 


1 10 


19 


58 50 




- 3 37 



19 55 13 
7 55 13 a.m. 



Example 2. 1895, Aug. 23, p.m., at sea, in latitude 9° 
37' N. Five observed altitudes of the O were taken at the 
times (by the watch) standing opposite : 

Readings of sextant © : 



21° 33' 30" 


2 42 58 


21° 28' 10" 


2 43 21.5 


21° 24' 00" 


2 43 39 


21° 19' 10" 


2 43 59.5 


21° 14' 30" 


2 44 18.5 



Index error, ; height of eye, 18 feet; correction of watch by 
chronometer, — 5 hours 16 minutes 49.3 seconds. Required 
the local time. 

Note. Since the determination of the time at sea requires that the 
altitude of the object should be taken -with great accuracy, a single 
observation for this purpose is seldom considered as sufficient ; it is, 
therefore, usual to take a set of altitudes and to employ the mean of the 



400 



NAVIGATION 



whole, taking the mean of the corresponding times by the watch as the 
estimated time. 

The mean of the five observed altitudes is found by dividing their 
sum by 5, and the mean time (by a watch) of these observations is found 
in a similar manner. 



Readings of sextant 

h. ra. s. 



21° 33' 30' 
21° 28' 10' 
21° 24' 00' 
21° 19' 10' 
21° 14' 30' 
119' 20' 
21° 23' 52' 



2 42 58 

2 43 21.5 

2 43 39 

2 43 59.5 

2 44 18.5 

218 16.5 

2 43 39.3 



We are to proceed, therefore, as if the 
observed altitude of the sun's lower limb 
was 21° 23' 52" and the corresponding 
time per watch 2 h. 43 m. 39.3 s.. p.m. 







2 43 39.3 






- 5 16 49.3 






9 26 50 a.m. 


Aug. 


22, 


21 26 50 G.M.T. 
2 33 10 = 2.55 h. 







'sdec, Aug. 23, Oh. 
11° 27' 14.7" N. 
+ 2' 9.9" 



11° 29 
90° 



24.6" N. 



p = 78° 30' 35' 



Semidiam. 

Parallax 

Dip 

Refraction 

Correction 

Eq. of time. 

m. s. 

2 32.07 

1.64 

2 33.71 



+ 15' 52' 
+ 8' 

- 4' 9' 

- 2' 28' 



= + 9' 23" 
Diff. for 1 h. 

s. 

- 0.642 
1.284 2 
0.321 0.5 
0.032 0.05 
1.637 



The Greenwich mean time is found to 
be Aug. 22 d. 21 h. 26 m. 50 s., which is 
2 h. 33 m. 10 s. before noon, Aug. 23. 



Diff. for 1 h. 
Diff. for 2 h. = 
Diff. for 0.5 h. = 
Diff. for 0.05 h. = 
Diff. for 2.55 h.= 



50.94' 
101.88' 

25.47' 

2.55' 

129.90' 

2' 9.9' 



L = 9° 37' 00' 
p = 78° 30' 35' 
h= 21° 33' 15' 



2 S = 109° 40' 50' 
S= 54° 50' 25' 
R= 33° 17' 10' 



log cos S= 9.76031 

log sin R= 9.73943 

log sec L= 0.00615 

log esc p - 0.00879 

2) 19.51468 

logsin£*= 9.75734 

£* = 34°53'3" 

t = 69° 46' 6" 



.-. t = 4 39 4 
Equation of time = + 2 34 
Local mean time = 4 41 38 p.m. 



LATITUDE BY MERIDIAN OBSERVATION 401 

SECTION XXVIII 
LATITUDE BY MERIDIAN OBSERVATION OF A HEAVENLY BODY 

Latitude is the angle between the direction of gravity and 
the plane of the equator. Hence, to find the latitude we have 
to find the distance of the zenith from the equator ; that is, 
the declination of the zenith. 

Let HZH' (Fig. 41) represent the meridian of the observer, 
P the nearest pole, Z the zenith, HH' the horizon, EQ the 
equator, and S the heavenly body on the meridian. 




Fig. 41 



QS = d — the declination of the heavenly body. 

HS = k = the true altitude of the heavenly body. 

SZ = z — 90° — h — the zenith distance of the heavenly 
body. The zenith distance is named N. or S. according as 
the zenith is north or south of the heavenly body. 

PQ = ZH' (each being equal to 90°) ; taking away ZP, we 
have PH ' = QZ ; that is, the altitude of the pole is equal to 
the latitude of the observer. 

QZ = SZ + QS, or L = z + d. 

If the heavenly body is at S', 

QZ = S'Z - QS', or L = z - d. 



402 NAVIGATION 

If the heavenly body is at S", 

QZ = QS" - ZS", or L — d — s. 
If the heavenly body is at S'", 

QZ = QS'" - ZS'", or L = d ' - z. 

Hence, to find the latitude of a place by meridian observa- 
tion of a heavenly body : 

If the declination and zenith distance of the body are of the 
same name, take their arithmetical sum ; if of different names, 
take their arithmetical difference and give it the name (N. or S.) 
of the greater. 

That is, the algebraic sum of the declination and the zenith 
distance is the latitude, and the latitude is of the name of the 
greater. 

The following method may also be used to find the latitude 
from the meridian altitude of a heavenly body : 

(1) The meridian altitude of the sun is found to be 25° from 
the south point of the horizon, that is, 65° south of the zenith, 
and the Nautical Almanac gives its declination on that day as 
15° S. Find the latitude of the observer. 

Solution. If the observer was in 15° S. latitude, the sun would be in 
the zenith ; as the sun is 65° south of the zenith, the observer must be 
65° to the north of 15° S ., or in latitude 50° N. 

(2) If the meridian altitude of the sun is 10° from the north 
point of the horizon, that is, 80° north of the zenith, and the 
Nautical Almanac gives its declination at the time as 20° N., 
what is the latitude of the observer? 

Solution. If the observer was in 20° N. latitude, the sun would be in 
the zenith ; as the sun is 80° north of the zenith, the observer must be 
80° to the south of 20° N., or in latitude 60° S. 

(3) The sun's meridian altitude is 60° from the north point 
of the horizon, and the Nautical Almanac gives the sun's 
declination as 10° N. Find the latitude of the observer. 



LATITUDE BY MERIDIAN OBSERVATION 



40; 



Solution. If the observer was in 10° N. latitude, the sun would be 
in the zenith ; as the sun is 00° — 60° = 30° north of the zenith, the 
observer must be 30° to the south of 10° N., or in latitude 20° S. 

(4) The declination of a star is 34° X., and its altitude when 
on the meridian is 56° 27' from the south point of the horizon. 
Find the latitude of the observer. 

Solution. If the observer was in 34° S. latitude, the star would be 
in the zenith ; as the star when on the meridian is 90° — 56° 27' = 33° 33' 
south of the zenith, the observer must be 33° 33' to the north of 34° N., 
or in latitude 67° 33' N. 

(a) Latitude by meridian altitude of the sun. 

The common mode at sea of measuring a meridian altitude 
of the sun is to commence observing the altitude 20 or 30 min- 
utes before noon, repeating the operation until the highest 
altitude is attained; soon after which the sun, as seen through 
the sight tube of the instrument, begins to dip or descend below 
the line of the horizon. 

Example. 1895, Aug. 11, in longitude 92° 12' E., by account, 
the observed meridian altitude of the sun's lower limb was 
42° 42' 10" N.; index error, -2' 50"; height of the eye, 17 
feet. Find the latitude. 

Since the sun bears north, the zenith distance is south. 

d. h. m. s. 

Ship date = Aug. 110 

Longitude 92° 12' E. = 8 48 

Greenwich date = Aug. 10 17 51 12 

-44.51" 
x6.15 
273.74" 
= 4' 34" 



Reading of sex- 
tant = 42° 42' 10" 


r Index error = — 2' 50' 
Semidiam. = + 15' 50' 


Cor. = +8' 2"* 


Dip =- 4' 2' 
Kefraction = — 1' 8' 


True 


Parallax = + 7 y 
■ Correction = + 8' 2' 






s dec. A us. 

= i5° i7' r 

+ 4' 34' 


11 

'N. 


d 
z 


= 15° 21' 41" N. 
= 47° 9/ 48" S. 


L 


= 31° 48' V 


' S. 



alt. ©: 



42° 50' 12' 
90° 



z = 47° 9'48"S. 



404 



NAVIGATION 



EXERCISE XIII 
From the following data find the latitude : 





Civil Date 


L.OXGITUDE 


Obs. Merid. Alt. 


Index Cor. 


Eye] 


1 


1895, Jan. 


1 


102° 41' w. 


59° 59' 50" S. 


+ 0' 50" 


■^i 


2 


1895, Feb. 


1 


78°14'E. 


78° 4'10"S. 


+ 0' 55" 


12 


3 


1895, Mar. 


20 


173° 18' W. 


89° 37' 0"N. 


+ 4' 32" 


18 


4 


1895, April 


1 


87°42'W. 


48° 42' 30" S. 


+ 1' 42" 


18 


5 


1895, Sept. 


1 


97°42 / E. 


51° 4'50"S. 


-6' 0" 


23 


6 


1895, Aug. 


26 


92° 3'E. 


35° 35' 20" N. 


+ 2' 17" 


12 


7 


1895, May 


16 


45° 26' W. 


86° 34' 20" N. 


+ 4' 16" 


15 


8 


1895, Mar. 


20 


174° 0'W. 


89° 56' 10" N. 


--T 15" 


15 


9 


1895, June 


1 


44°40'E. 


72° 14' 10" N. 


+ 3' 45" 


22 


10 


1895, Dec. 


1 


67°56 , E. 


18° 48' 10" S. 


-3' 6" 


18 


11 


1895, Sept. 


23 


57°45 , E. 


84° 10' 50" N. 


- V 36" 


16 


12 


1895, Sept. 


23 


119° 54' E. 


83° 46' 0"S. 


- 5' 30" 


18 


13 


1895, Nov. 


21 


70°20 , E. 


80° 20' 0"N. 


- 2' 50" 


20 


14 


1895, Dec. 


31 


123° 45' W. 


67° 8'10"S. 


+ 0' 9" 


13 


15 


1895, Oct. 


20 


150° 25' W. 


49° 58' 50" N. 


+ 1' 10" 


19 


16 


1895, June 


1 


96°17 / E. 


75° 38' 15" N. 
Obs. Merid. alt. 


+ 0' 27" 


26 


17 


1895, June 


25 


59° 15' E. 


60° 23' 15" N. 


+ 2' 21" 


30 



(b) Latitude by meridian altitude of a fixed star. 

Since the declination of a fixed star changes very slowly, it 
is sensibly the same for several successive days ; hence, the 
Greenwich date and the longitude are not required. 

Example. 1895, Dec. 24, the observed meridian altitude of 
the star Procyon (declination 5° 29' 34" K) was 52° 51' 50" N. ; 
index error, — 49" ; height of eye, 21 feet. Find the latitude. 

Beading of sextant = 52° 51' 50" N. Index error = - 0' 49" 



Correction 


= -6' 2" 


Dip = - 4' 29" 


True altitude 


■= 52° 45' 48" N. 


Refraction = - 0' 44" 




90° 


Correction = - 6' 2" 


Zenith distance 


= 37° 14' 12" S. 




Declination 


= 5° 29' 34" N. 




Latitude 


= 31° 44' 38" S. 





LATITUDE BY MERIDIAN OBSERVATION 



405 



Declinations of Fixed Stars 



Date 


Stab 


Declination 


1895, Jan. 29 


Aldebaran 


16° 18' 2" N. 


1895, Feb. 8 


Sirius 


16° 34' 20" S. 


1895, Feb. 18 


Procyon 


5° 29' 39" N. 


1895, Mar. 20 


Arcturus 


19° 43' 23" N. 


1895, Mar. 30 


Spica 


10° 37' 4"S. 


1895, April 9 


Sirius 


16° 34' 24" S. 


1895, July 8 


Antares 


26° 12' 12" S. 


1895, Aug. 17 


Altair 


8° 35' 34" X. 


1895, Aug. 17 


Ceutauri 


59° 52' 29" S. 


1895, Sept. 6 


Arcturus 


19° 43' 37" N. 


1895, Oct. 6 


Markab 


14° 38' 49" N. 


1895, Nov. 5 


Foraalhaut 


30° 10' 32" S. 


1895, Dec. 4 


a Arietis 


22° 58' 26" N. 


1895, Dec. 24 


Procyon 


5° 29' 34" N. 



EXERCISE XIV 



Using the declinations of fixed stars given in the above 
table, find the latitude: 



Civil Date 


Star 


Obs. Mebid. Alt. 


Index Cor. 


Eye 


1 1895, 


Jan. 29 


Aldebaran 


52° 36' 0"S. 


- 0' 23" 


20 


2 1895, 


Feb. 18 


Procvon 


77° 18' 10" S. 


+ 0' 19" 


16 


3 1895, 


Mar. 20 


Arcturus 


36° 10' 20" N. 


+ 2' 42" 


20 


4j 1895, 


Aug. 17 


Altair 


66° 51' 10" N. 


+ 0' 58" 


13 


5 


1895, 


Nov. 5 


Fomalhaut 


59° 40' 0"N. 


+ 1' 12" 


23 


6 


1895, 


Sept. 6 


Arcturus 


86° 35' 50" N. 


- V 10" 


12 


7 


1895, 


Oct. 6 


Markab 


54° 10' 15" S. 





13 


8 


1895, 


Aug. 17 


/3 Centauri 


59° 47' 13" S. 





25 


9 


1895, 


Dec. 4 


a Arietis 


60° 29' 50" S. 


- 2' 10" 


18 


10 


1895, 


Feb. 8 


Sirius 


37° 50' 20" S. 


+ 1' 4" 


19 


11 


1895, 


April 9 


Sirius 


61° 3'50"N. 





16 


12 


1895, 


Mar. 30 


Spica 


52° 14' 0"N. 





19 


13 


1895, 


July 8 


Antares 


70° 10' 30" N. 





21 



406 NAVIGATION 

SECTION XXIX 

LATITUDE BY EX-MERIDIAN ALTITUDE OF THE SUN BY 
REDUCTION TO THE MERIDIAN 

If it happens that the meridian altitude of the sun cannot 
be observed owing to clouds or other causes, the altitude may 
be observed before or after noon and reduced to the meridian 
altitude. 

Towson's Method 

Towson's Tables for the reduction of ex-meridian altitudes 
furnish a simple practical solution of the problem under con- 
sideration. 

Corresponding to the nearest hour angle (Sect. XXVI, p. 397) 
and nearest declination (Sect. XXV, p. 394) of the sun at 
the time of observation, Towson's first table gives an Index 
Number and Augmentation I. The declination of the sun at 
noon is found by adding Augmentation I to the declination 
at the time of observation. 

Corresponding to the Index Number (given by the first 
table) and the nearest true altitude of the sun at the time of 
observation, Towson's second table gives Augmentation II. 
The true altitude of the sun at noon is found by adding Aug- 
mentation II to the true altitude at the time of observation. 

The declination and the altitude at noon being known, the 
latitude is found as in Sect. XXVIII, p. 401. 

The latitude may be found from these tables by ex-meridian 
altitudes of a star or planet, but sidereal time must be employed 
in finding the hour angle. 

If equal altitudes are observed before and after meridian 
passage, the hour angle may be found by taking half of the 
elapsed time ; or, if the two altitudes differ only a few minutes 
(of arc), their mean may be reduced by employing half the 
elapsed time as the hour angle. 



LATITUDE BY EX-MERIDIAN ALTITUDE 



407 



It should be observed that Towson's method is independent 
of the latitude by account. 

The maximum altitude in Towson's Tables is 74°, and the 
hour angle corresponding cannot exceed 22 minutes. The 
greatest declination is 23° 20'. 

Towson's method will be illustrated by a single example. 



Example. 1895, Dec. 2, p.m., at ship, in longitude 4° 39' W. ; 
observed altitude © south of observer was 23° 14' 10"; height 
of eye, 11 feet; time by watch, Dec. 2, hours 50 minutes 
58 seconds, which had been foimd to be 19 minutes 38 seconds 
fast on apparent time at ship ; the difference of longitude made 
to the west was 21.3' after the error on apparent time was 
determined. Required the latitude. 



Time by watch 
Watch fast 



Dec. 



d. h. m. b. 

2 50 58 

- 19 38 



Diff. long. 4f 

App. time at ship = Dec. 

Time from noon = 



2 31 20 
- 1 25 



2 29 55 
29 55 



App. time at ship =Dec. 2 29 55 
Longitude, 4° 39' W. = + 16 36 

Greenwich date =Dec. 2 46 31 

46 m. 31 s. = 0.78 h. 
O'sdec, Dec. 2 d. h. 

= 21° 58' 41.3" 22.23" 
17.3" 0.78 



Reduced declination 
Aug. (Towson, Table I) 
Meridian declination 



= 21° 58' 59' 
= + 10' 17' 
= 22° 9' 16' 



17.34" 



Reading of sextant 



Correction 



True altitude 

Aug. (Towson, Table II) 

Meridian altitude 

Zenith distance 
Meridian declination 
Latitude 



23° 14' 10' 



+ 10' 54' 



= 23° 


25' 


4" 


= +iu 


2" 


= 23° 


36' 


6" 


90° 






= 66° 


23' 


54" 


= 22° 


9' 


16" 



Semidiameter 


= + 16' 16" 


Dip 


= - 3' 15" 


Parallax ' 


= + 0' 8" 


Refraction 


= - 2' 15" 


Correction 


= + 10' 54" 



44° 14' 38" N. 



408 NAVIGATION 

SECTION XXX 

LONGITUDE BY CHRONOMETER FROM AN OBSERVED ALTITUDE 
OF THE SUN 

The difference of longitude of two places is equal to the 
difference of time reduced to arc. (Sect. XXII, p. 389.) 

Hence, the longitude of a place from Greenwich may be 
determined by taking the difference between the local mean 
time and the Greenwich mean time and reducing to arc. The 
longitude will be east or west according as the Greenwich time 
is less or greater than local time. 

The local mean time is found as in Sect. XXVII, p. 398. 
The Greenwich mean time is found by chronometer set to 
Greenwich time. (See Sect. XXIV, p. 392, also the example 
of Sect. XXIX, p. 407.) 

As an astronomical question the determination of longitude 
resolves itself into the determination of the difference of time 
reckoned at the two meridians at the same absolute instant. 

If we could note the exact moment the sun is on the 
meridian, it would then be 12 o'clock apparent local time, 
and by applying the equation of time to it we should have 
the mean local time. The difference between it and the time 
shown by the chronometer at the moment of observation 
would be the difference of longitude in time. But we cannot 
determine the instant that the sun attains his meridian alti- 
tude. It is true that in the tropics, where the sun has always 
a high meridian altitude and where its motion in altitude is 
sufficiently rapid for a good observation at any point of his 
course, we can mark it pretty clearly ; but in other latitudes, 
especially when the sun's meridian altitude is low, we cannot 
mark it within three or four minutes, and a minute of time 
represents 15' of longitude. Navigators never rely on this 
method to determine longitude. 



LONGITUDE BY CHRONOMETER 409 

The longitude of any place on the surface of the earth is 
ascertained as soon as we can discover the time at that place 
and the time at Greenwich at the same instant, since we have 
only to change the difference of the two times into degrees 
by multiplying by 15. 

In the morning or evening, when the sun is 15° or 20° high 
and is rising or falling rapidly, the navigator observes its alti- 
tude, and notes the time by chronometer. He then solves a 
spherical triangle, of which he has given the three sides, the 
altitude just observed, the latitude, and the sums polar distance. 
The latitude is really found at noon, as previously explained, 
and is u worked back " to the time of taking the morning 
observation. The polar distance of the sun is 90° ± its decli- 
nation, which declination is found in the Nautical Almanac. 

Example 1. 1895, Feb. 10, a.m., at ship, latitude 50° 48' N. ; 
observed altitude O. 9° 10' 50"; index correction, - 3' 18": 
height of eye, 18 feet; time by chronometer, Feb. 9 days 
9 hours 59 minutes 25 seconds, which was 37 minutes 58.8 
seconds fast for mean noon at Greenwich, Dec. 20, 1894, and 
on Jan. 10, 1895, was 34 minutes 12 seconds fast for mean 
noon at Greenwich; from the time of observation until noon 
the ship has sailed on a S.W. course (true) 34 miles. Required 
the longitude at noon. 

At sea the latitude by account is used, either brought for- 
ward from a preceding determination or carried backward 
from a subsequent determination. 



Dec. 20, chronometer fast 
Jan. 10, chronometer fast 
Change in 21 days 



= 37 58.8 




f 3)226.8 


= 84 12.0 
= 3 46.8 


21. 


7)75.6 

10.8 losing 


60 




324.0 30 


226.8 




4.3 0.4 

60)328.3 

5' 28.3" 



410 



NAVIGATION 



The change in 1 day is 226.8 seconds -f- 21 = 10.8 seconds ; the change 
in 30.4 days (the interval between noon Jan. 10 and Feb. 9 days 10 hours) 
is 30.4 x 10.8 seconds = 328.3 seconds = 5 minutes 28.3 seconds. 



Time by chronometer = 

Original error = 

Accumulated rate = 

Greenwich mean time = 

Time, after noon 9 h. 30 m. 41 s. = 
O's declination = 14° 39' 24.8" S. 
- 7' 38.1" 

14° 31' 46.7" 

90° 



d. h. m. s. 

Feb. 9 9 59 25 
-34 12 



9 9 25 

+ 5 



9 9 30 41 



9.51 h. 



p = 104° 31' 

Reading of sextant 

= 9° 10' 50 
Correction = + 3' 10 



+ 48.17' 
x 9.51 
458.10' 

7' 38.1' 



Semidiam. = +16' 15' 
Parallax = + 9' 



True alt. 0=9° 14' 
L= 50° 48' 0" 
p =104° 31' 47' 
h= 9° 14' / 

2 5=164° 33' 47' 
8 = 82° 16' 53' 
B= 73° 2' 53' 



Ind. er. = 
Dip = 
Ref. 



log cos 8 = 9.12810 
log sin R = 9.98071 
log sec L= 0.19926 
log cscp = 0.01411 
2 )19.32218 
logsin££= 9.66109 

t = 



Eq. 



Cor. 

of time. 



14 25.05 

+ 0.53 

14 25.58 



- 3' 18' 

-4' 9' 

- 5' 47' 
+ 3' 10' 



+ 0.056 
X9.51 



0.533 



27° 16' 24" 
54° 32' 48" 
3 h. 38 m. 11 s. Time before noon, Feb. 10. 



Feb. 9 20 21 49 local apparent astronomical time. 

+ 14 26 equation of time. 
Feb. 9 20 36 15 local mean astronomical time. 
Feb. 9 9 30 41 Greenwich mean time. 
11 5 34 difference of time. 
166° 23' 30" E. long, at time of observation. 
Longitude at sights = 166° 23' E. Latitude at sights = 50° 48' N. 

Change until noon = _ 38' W . Change until noon = 24' S. 

Longitude at noon = 165° 45' E. Latitude at noon = 50° 24' N. 



LONGITUDE BY CHRONOMETER 



411 



From the time of observation until noon the ship sailed 8. W. 34 miles. 
The difference of latitude and the difference of longitude corresponding 
to this course and distance are found by Middle Latitude Sailing to be 
24' S. and 38' W. 

At sea it is usual to reduce longitudes obtained from day observations 
to noon by allowing for the run of the ship in the interval. 

Example 2. 1895, Nov. 8, p.m., at sea, in latitude 30' S.; 
five observed altitudes of the ©_ were taken at the times (by 
watch) standing opposite : 

Headings of sextant 0: 24° 51' 50" 

48' 50" 
44' 40" 
40' 20" 
37' 0" 

Index error, + 30"; height of eye, 18 feet ; correction of 
watch by chronometer, — 5 hours 12 minutes 15.9 seconds. 
Required the longitude. 



)l. 


m. s. 


11 


8 3 




16 




33 




49 




64.5 





h. in. s. 












24° 51' 50" 


11 8 3 


0's declination 








48' 50" 


1G 


16° 35' 


43.2" S. 




- 43.60" 




44' 40" 


33 

49 


+ 4' 


19.0" 




x 5.94 

258.9b" 




40' 20" 


16° 40' 


2.2" S. 




37' 0" 


64.5 


90° 






= 4' 19.0" 




222' 40" 


165.5 


p = 73° 19 


58" 






24° 44' 32" 


11 8 33.1 




Semidiameter 


= + 16' 


11" 




-5 12 15.9 




Parallax 


= + 


8" 




5 56 17.2 




Index error 


= + 


30" 




= 5.94 h. p.m. 




Dip 


= - 4' 


9" 


Reading of se 


xtant = 24° 44' 
= +10' 


32" 
34" 


Refraction 
Correction 


= - T 


6" 


Correction 


= + 10' 


34" 


True altitude = 24° 55' 


6" 








L = 0° 30' 


0" log cos S - 


= 9.81365 








r. = 73° 19' 


58" log sin R -. 


= 9.61701 


1 1 = 32° 


2' 14". 




h = 24° 55' 


6" log sec L - 


= 0.00002 


£ = 64° 


4' 28". 




2 S = 98° 45' 


4" log esc p = 


= 0.01864 


.-. t = 4 h 


16 m. 18 


s. 


S = 49° 22' 


32" 2 ; 


19.44932 


Time after noon, Nov. 


8. 


R = 24° 27' 


26" log sin $ t - 


= 9.72466 











412 



NAVIGATION 



Eq. of time. 



169.10 
-1.07 
16 8.03 



- 0.180 
x 5.94 



1.07 



Nov. 8 4 16 18 local apparent astronomical time. 

16 8 equation of time. 
Nov. 8 4 10 local mean astronomical time. 
Nov. 8 5 56 17 Greenwich mean time. 

7 difference of time. 

' W. longitude. 



1 56 
29° V 45 



Problem foe, the " Flying Dutchman"* 

In what way could a navigator who has no knowledge 
of his position and has lost all record of time ascertain not 
only his correct latitude and longitude 
but recover the day and date ? 

Let H'ZH (Fig. 42) represent the 
meridian, P the pole, Z the zenith, 
EQ the equator, TT' any circumpolar 
star, and H'H the horizon. 

(1) Observe altitude of a circum- 
polar star at upper and lower culmi- 
nation. One half the sum of these 
altitudes is equal to the altitude of 
the pole, which in turn equals the latitude of the place. 

(2) Heave-to the ship so as to keep her on the same parallel 
of latitude until noon of the next day. 

(3) At noon observe the meridian altitude of the sun, which 
subtracted from 90° gives zenith distance. 

(4) Knowing zenith distance and latitude we can find sun's 
declination. 

(5) Look in Nautical Almanac to see on what day the sun 
has this declination. 




FlG. 42 



* Lecky's Wrinkles in Practical Navigation. 



LONGITUDE BY CHRONOMETER 413 

Note. There are, however, two difficulties in carrying out these 
directions in actual practice : 

(1) As the observer is not supposed to know whether the sun is moving 
north or south, it would be necessary to wait and take two meridian 
altitudes to determine on which side of the solstice the sun is. 

(2) As the observer's longitude is unknown and as the elements in the 
Nautical Almanac are computed for noon Greenwich date and not for 
noon at ship, he would still be uncertain as to the exact date. 

EXERCISE XV 

1. 1895, Oct. 19, a.m., at sea, in latitude 33° 27' S. ; the 
observed altitude O, 28° 22' 30"; index correction, +30"; 
height of eye, 18 feet ; Greenwich mean time by chronometer, 
Oct. 18 d. 18 h. 28 m. 38 s. Required the longitude. 

2. 1895, Oct. 20, a.m., at sea, in latitude 31° 40' S. ; the 
observed altitude 0, 35° 16' 10"; index correction, -f30"; 
height of eye, 18 feet ; Greenwich mean time by chronometer, 
Oct. 19 d. 19 h. 11 m. 24 s. Required the longitude. 

3. 1895, Oct. 20, p.m., at sea, in latitude 30° 55' S. ; the 
observed altitude 0, 21° 42' 30"; index correction, +29"; 
height of eye, 18 feet ; Greenwich mean time by chronometer, 
Oct. 20 d. 3 h. 35 m. 40 s. Required the longitude. 

4. 1895, Oct. 21, a.m., at sea, in latitude 29° 35' S. ; the 
observed altitude O, 24° 26' 42"; index correction, +29"; 
height of eye, 18 feet ; Greenwich mean time by chronometer, 
Oct. 20 d. 18 h. 30 m. 39 s. Required the longitude. 

5. 1895, Jan. 29, p.m., at ship, latitude 42° 26' X. ; observed 
altitude ©., 13° 40'; index correction, — 1'8"; height of eye, 
16 feet ; time by chronometer, 29 d. 6 h. 48 m. 40 s., which 
was slow 11 m. 22.3 s. for mean noon at Greenwich, Dec. 1, 
1894, and on Jan. 1, 1895, was 8 m. 7 s. slow for Greenwich 
mean noon. Required the longitude. 

6. 1895, March 31, a.m., at ship, latitude 26° 9' X. ; observed 
altitude ©. 29° 10' 20"; height of eye, 26 feet; time by 



414 NAVIGATION 

chronometer, 31 d. h. 4 m. 50 s., which, was 58 m. 58 s. fast 
for mean noon at Greenwich, Nov. 20, 1894, and on Dec. 31, 
1894, was 1 h. 2 m. 55.8 s. fast for mean time at Greenwich. 
Required the longitude. 

7. 1895, May 22, a.m., at ship, latitude 43° 25' N. ; observed 
altitude O., 32° 8'; index correction, -f 1' 28"; height of eye, 
15 feet; time by chronometer, 21 d. 21 h. 6 m. 10 s., which 
was slow 12.6 s. for mean noon at Greenwich, Feb. 24, and 
on April 1 was 2 m. 45 s. fast for mean noon at Greenwich. 
Required the longitude. 

8. 1895, Aug. 24, a.m., at ship, latitude at noon, 37° 59' K ; 
observed altitude O, 37° 13' 30"; index correction, + 2' 44"; 
height of eye, 18 feet ; time by chronometer, Aug. 23 d. 18 h. 
13 m. 24 s., which was 1 m. 5 s. fast for mean noon at Green- 
wich, Aug. 1, and on Aug. 10 was m. 42 s. slow for mean 
time at Greenwich; course (true) since observation, N.N.W.; 
distance, 22.4 miles. Required the longitude at noon. 

9. 1895, Jan. 29, p.m., at ship, latitude 28° 45' ST. ; 
observed altitude O, 17° 46' 30"; index correction, - 3' 18"; 
height of eye, 16 feet ; time by chronometer, Jan. 28 d. 16 h. 
31 m. 30 s., which was 1 m. 16.5 s. fast for Greenwich mean 
noon, Dec. 17, 1894, and on Jan. 1, 1895, was 1 m. 3 s. slow 
for Greenwich mean time ; course (true) since noon, N.W. by 
W. ; distance, 20 miles. Required the longitude at the time 
of observation, and also at noon. 

10. 1895, Aug. 31, p.m., at ship, latitude 0°; observed 
altitude O, 45° 5' 30"; index correction, -2' 4"; height of 
eye, 15 feet ; time by chronometer, Aug. 31 d. 9 h. 11 m. 
28 s., which was 5 m. 20 s. fast for mean noon at Greenwich, 
April 15, and on June 16 was fast 2 m. 43 s. on mean time at 
Greenwich. Required the longitude. 

11. 1895, April 15, a.m., at ship, latitude 48° 52' K ; 
observed altitude O, 22° 18'; index correction, —3' 54"; 



LONGITUDE BY CHRONOMETER 415 

height of eye, 17 feet ; time by chronometer, April 14 d. 22 fa. 
30 m. 42 s., which was m. 4 s. slow for mean noon at Green- 
wich, Jan. 1, and on Jan. 12 was fast m. 2 s. Required 
the longitude. 

12. 1895, Aug. 28, p.m., at ship, latitude 5°S.; observed 
altitude O, 38° ; index correction, + 5' 27" ; height of eye, 
21 feet ; time by chronometer, Aug. 27 d. 22 h. 20 m. 30 s., 
which was 10 m. s. slow for mean noon at Greenwich, 
Feb. 19, and on May 30 was 2 m. 20 s. slow on mean noon 
at Greenwich. Required the longitude. 

13. 1895, Sept. 22, a.m., at ship, on the equator, observed 
altitude ©, 17° 20' 40"; index correction, —V 9"; height of 
eye, 20 feet ; time by chronometer, Sept. 22 d. 4 h. 59 m. 16 s., 
which was 15 s. slow for Greenwich mean noon, April 30, 
and on June 1 was 10.6 s. fast for mean time at Greenwich. 
Required the longitude. 

14. 1895, Aug. 5, a.m., at ship, latitude at noon, 30° 30' 
N". ; observed altitude ©, 35° 6'; height of eye, 15 feet; time 
by chronometer, Aug. 5 d. 8 h. 39 m. 22 s., which was fast 
29 m. 32.4 s. on Greenwich mean noon, July 8, and on July 20 
was fast 30 m. 0. s. for mean noon at Greenwich ; course (true) 
till noon, W. ; distance, 48 miles. Required the longitude at 
noon. 

15. 1895, Nov. 12, a.m., at sea, in latitude 7° 10' NT. ; four 
observed altitudes of the Q were taken at the times (by 
watch) standing opposite: 



Obs. alt. : 


21° 8' 40" 




11' 50" 




14' 50" 




17' 30" 



h. m. 


s. 


2 55 


48 


56 





56 


13 


56 


26.5 



Index correction, -f31"; height of eye, 18 feet; correction 
of watch by chronometer, —5 h. 12 m. 2.1 s. Required the 
longitude. 



416 NAVIGATION 



16. 1895, Nov. 13, a.m. 


, at 


sea, in latitude 9° 30' N. ; 


five observed altitudes of 


the 


o 


were taken at the times 


(by watch) standing opposite: 




h. ni. s. 


Obs. alt. © : 18° 58' 40" 






2 59 2 


19° 1'20" 






13 


3' 30" 






28 


V 30" 






45 


11' 0" 






57.5 


Index correction, -f 32" ; 


height 


of eye, 18 feet ; correction 


of watch by chronometer, - 


-5h 


. 11 m. 58.4 s.. Required the 


longitude. 








17. 1895, Nov. 17, a.m., 


at 


sea 


, in latitude 15° 35' N. ; 


five observed altitudes of 


the 


o 


were taken at the times 


(by watch) standing opposite : 




h. m. s. 


Obs. alt. O : 23° 56' 0" 






4 12 31 


24° 0' 0" 






46 


4' 0" 






13 2.5 


6' 10" 






14 


10' 0" 






28.5 



Index correction, +31"; height of eye, 18 feet ; correction 
of watch by chronometer, — 5 h. 11 m. 43.6 s. Required the 
longitude. 

18. 1895, Nov. 18, a.m., at sea, in latitude 16° 25' N.; 
five observed altitudes of the O were taken at the times 
(by watch) standing opposite: 

Obs. alt. O : 18° 13' 30" 
16' 10" 
19' 20" 
22' 30" 
25' 30" 
Index correction, + 32" ; heig 
of watch by chronometer, — 5 h. 11 m. 39.9 s. Required the 
longitude. 



h. 


m. s. 


3 


52 42 




53.5 




53 6.5 




23 




38 


f e 


ye, 18 feet ; correction 



1). 


m. 


8. 


6 


27 


14 

29.5 

49 




28 


5 

23 



DEVIATION BY TIME AZIMUTHS 417 

19. 1895, Dec. 4, a.m., at sea, in latitude 36° 10' K ; 
five observed altitudes of the ©_ were taken at the times 
(by watch) standing opposite: 

Obs. alt. O : 13° 0' 30" 

, 3' 10" 

5' 40" 

8' 50" 

12' 0" 

Index correction, + 32" ; height of eye, 18 feet ; correction 
of watch by chronometer, — 5 h. 10 m. 47.1 s. Required the 
longitude. 

20. 1895, Dec 4, p.m., at sea, in latitude 36° 36' N.; 
four observed altitudes of the O. were taken at the times 
(by watch) standing opposite: 

Obs. alt. O : 16° 31' 10" 1 7 26.5 
30' 0" 35.5 

28' 30" 46 

27' 20" 56.5 

Index correction, +30"; height of eye, 18 feet ; correction 
of watch by chronometer, — 5 h. 10 m. 46.1 s. Required the 
longitude. 

SECTION XXXI 

DEVIATION BY TIME AZIMUTHS 

The Azimuth of a heavenly body is the arc of the horizon 
included between the north point and the south point and the 
vertical circle passing through the body. It is named by pre- 
fixing the letter N. or S., which indicates the point from which 
it is reckoned, and affixing the letter E. or W., which indicates 
upon which side of the meridian of the observer the heavenly 
body is situated. 



418 NAVIGATION 

The True Azimuth is reckoned from the true north or 
south point. 

The Compass Azimuth is reckoned from the magnetic north 
or south point. 

When a heavenly body is on the horizon the arc of the 
horizon between the body and the east or west point is called 
the Amplitude of the body. 

To find the compass azimuth of a heavenly body : 

The compass azimuth is observed directly by means of the 
Azimuth Compass. 

This instrument differs from the ordinary compass chiefly in 
having two sight vanes hinged to opposite points of the bowl. 

The standard compass is commonly also an azimuth compass. 

To observe the compass azimuth, the observer turns the 
compass so that by looking through the narrow slit in one 
of the sight vanes he can see the object, or the image of it, 
reflected by a small mirror, bisected by the vertical wire of 
the other vane. A spring is then touched which holds the 
card in position until the observer reads off the division of the 
card, apparently in the prolongation of the wire of the vane. 

To find the true azimuth of a heavenly body : 

First Method ; by computation. The necessary data for time 
azimuths are : the declination (d), the hour angle (t), and the 
latitude (L). Let a be the true azimuth. 

Formulas (A) and (C) of Spherical Trigonometry, Sect. 
LXXI, p. 193, apply. 

tan m = cot d cos t. 

tan a == sec(Z> + m) sin m tan t. 

Note. In these formulas m represents the polar distance of the foot 
of the perpendicular let fall from the body upon the meridian. 

Second Method; by inspection. Time azimuth tables have 
been prepared which give the true azimuth of the sun (and 



DEVIATION BY TIME AZIMUTHS 419 

other heavenly bodies whose declinations lie between 23° X. 
and 23° S.) corresponding to the given latitude, declination, 
and apparent local time. This is the best practical method of 
finding true azimuths. 

To find the total error of the compass : 

The total error of the compass is equal to the difference 
between the true azimuth and the compass azimuth of a 
heavenly body when they are both reckoned from the points 
of like name toward the east or toward the west ; it is equal 
to the sum of the true azimuth and the compass azimuth 
when one is reckoned toward the east and the other toward 
the west. 

To name the total error of the compass, let the observer 
look from the centre of the compass toward the two azimuths ; 
then the error is east or west according as the true azimuth 
falls to the right or left of the compass azimuth. 

To find the deviation : 

If the total error of the compass and the variation (given 
by the chart) is of the same name (E. or W.), their differ- 
ence is the deviation ; if of different names, their sum is the 
deviation. 

When the error exceeds the variation, the deviation is of 
the same name as the error ; but when the error is less than 
the variation, the deviation is of contrary name to the error. 

The deviation is thus determined for the point of the com- 
pass toward which the ship's head lies at the moment of 
observation. By swinging the ship and repeating the obser- 
vations the deviation may be found for as many points of the 
compass as may be desired. Simple interpolation may be 
employed in finding the deviation for a point situated between 
two points for which the deviations are known. 





Exercise I 


:. Page 2 


5 * . 13 » 


; re* 


. 1 1 

> T5 


jr; 


&*• 


; 112° 30' 


; 168° 


45'; 


; 84 





0.0002909. 






8. 

9. 
10. 
11. 


69.166 miles. 

57 feet 3.55 inches. 

3 hours 49 minutes 11 seconds. 

9 feet 2 inches. 



ANSWERS 

PLANE TRIGONOMETRY 



1. itf; }*; 

2. 120°; 135 c 

3. 0.017453; 

4. 206,265 ,/ . 

5. far; f jr. 

6. 11° 27' 33" 

7. 14° 27' 28". 12. T | 3 seconds. 

Exercise II. Page 5 

1 . sin B = - ; cos B = - ; tan jB = - ; cot B = r ; sec B = - ; esc B = =- . 
c c a o a o 

3. (i) sin = |, cos = i , tan = f , cot = f , sec = f, csc = f ; 
(ii) sin = T 5 3 , cos = if, tan = y 5 ^, cot = - 1 /, sec = if, esc = - 1 / ; 

(iii) sin = T 8 T , cos = |f , tan = T 8 5 , cot = - 1 /, sec = Jf, esc = *£ ; 

(iv) sin = T 9 T , cos = £f , tan = T %, cot = -\°, sec = |£, esc = - 4 ^- ; 

(v) sin = ff, cos = ff, tan = §{f, cot = |f, sec = ff , csc = f f ; 
(vi) sin = Hf, cos = iff, tan = iif , cot = iff, 
sec = iff, esc = Jf |. 

4. The required condition is that a 2 + fr 2 = c 2 . It is. 

m ,„ . 2 mn m 2 — n 2 2 mn 

5. (l) sin = — , cos = — -j tan = — > 



COS 


m 2 - n 2 


m 2 + n 2 


sec 


m 2 + n 2 


m 2 — n 2 


cos 


x 2 -y 2 

~ x 2 + y 2 ' 


sec 


Z 2 + i/ 2 



w 2 - n 2 m? + n 2 m 2 + n 2 

cot = > sec = » esc = 

2 mn m 2 — n 2 2 mn 



.... . 2xy x 2 -y 2 2xy 

(n) sm = j cos = — , tan = 

v x 2 -\-y 2 x 2 + y 2 x 2 -y 2 

x* — 2/2 x 2 + y 2 x 2 + y 2 

cot = > sec = , esc = 

2xy x 2 -y 2 2 xy 



PLANE TRIGONOMETRY 



(iii) sin 
(iv) sin 

cot 

7. In (iii) 

8. sin .4 = 
cot A - 

9. sin A : 
cot A - 

10. sin A -- 
cot A -. 



V 



tan = — , cot = - , 
s p 



V 
sec = — 

Q 



qr 

_ pv 



— ± , cos = - 
s p 

ms mpv ns 

cos = > tan = — , 

nqr pv 

nqr qr 

sec = — 1 —< ) esc = -^— . 
ns mpv ms 

p 2 q 2 + q-s 2 = p 2 s 2 ; in (iv) m 2 n 2 s 2 + m 2 p 2 v 2 = n 2 q 2 r 2 . 

-- T \\ = cosB; cos^ = iff = ~ : 

- 143 — 



CSC 



if| = sinJ5; tan^=^ 3 

££ = tan 5 ; sec A = Iff = esc B ; esc A = ±£g- = see B. 

f f = cos B ; cos J. = 2 2 e\ = sin -B J tan ^ = *$£ — cot 5 ; 

- 2 ^ = tan JB ; sec ^1 = ^ = esc B ; esc J. = § § £ = sec i?. 

1 1| = cos B ; cos ^1 = T 9 g 5 3 = sin B ; tan A = ^- = cotB; 

t <ul = tan B j sec J. = V/ = csc jB ; esc 4 = iff = sec B. 



= cosB; 



= cotB 




v2pq 

cos A = — = 

P + q 



— sin B 



cot A = 



V2 



pq 



V«2" 



= tan B 



/p 2 + q'< 
esc J. = ^ = sec B. 

Vp 2 

= sin B 




= sec 5. 



cot A = < ?—M = tan 2? ; 

esc ^1 = - — - = sec B. 
p-q 



14. sin A 
sec .A 

15. sin A 
sec .4 



cot A = I; 



: f- ; cos A = £ V& ; 

: | Vb .; esc ^4 = f . 



tan ^4 = 2 ; 

tan ^. = | V5 ; cot A = \ VI ; 



ANSWERS 3 

16. sin A = i(5+V7); cosA = \(b-V7) ; 
tan A = I (16 + 5 V7) ; cot ^i = £ (16 - 5 V7) ; 
sec A = f (5 + V7) ; esc A = f (5 - V7). 

17. sin^ = }(V31 + l) ; cos^l = |(V31-1); 
tan A = T \ (16 -f V31) ; cot A = ^ (16 - V31) ; 
sec A = T 4 5 ( V31 + 1) ; esc A = T %(^S1 - 1). 

18. a = 12.3. 20. a = 9. 22. c = 40. 

19. b = 1.54. 21. b = 68. 23. c = 229.62. 

24. Construct a rt. A with legs equal to 3 and 2, respectively ; then con- 
struct a similar A with hypotenuse equal to 6. 
28. a = 1.5 miles ; b = 2 miles. 
30. a = 0.342, b = 0.940 ; a = 1.368, 6 = 3.760. 31. 142.926 yards. 

Exercise III. Page 9 

5. Through A (Fig. 3) draw a tangent, and take AT equal to 3; the 

angle A T is the required angle. 

6. From (Fig. 3) as a centre, with a radius equal to 2, describe an 

arc cutting at S the tangent drawn through J5; the angle A OS is 
the required angle. 

7. In Fig. 3, take OM equal to £, and erect MP _L OA, intersecting the 

circumference at P ; the angle POM is the required angle. 

8. Since sinx = cosx, OM = PM (Fig. 3), and x = 45°; hence, con- 

struct x equal to 45°. 

9. Construct a rt. A with one leg equal to twice the other ; the angle 

opposite the longer leg is the required angle. 

10. Divide OA (Fig. 3) into four equal parts ; at the first point of divi- 
sion from O erect a perpendicular meeting the circumference at 
some point P. Draw OP ; the angle A OP is the required angle. 

12. x = 18°. 21. rsinx. 22. a = mc; b = nc. 

Exercise IV. Page 12 

1 . cos 60° ; sin 45° ; cot 1° ; tan 75 ° ; 

sec 71° 50' ; sin 52° 36 r ; tan 7° 41' ; sec 35° 14'. 

2. cos 30°; sin 15°; cot 33°; tan 6°; 

. sec.20° 58' ; sin 4° 21' ; tan 0° V ; sec 44° 59'. 



PLANE TRIGONOMETRY 



3. *V§. 


6. 30°. 


9. 22° 30'. 


4. 45°. 


7. 90°. 


10. 18°. 


5. 30°. 


8. 60°. 


11. 10°. 



12. W 



n + 1 



Exercise VI. Page 16 

1. cosA= T 5 j; tanJ.=-y-'; cotA = T \; secA=±£; csc^.=if. 

2. cos A =0.6; tan4 = 1.3333; cot A =0.75; sec^i = 1.6667; csc^l = 1.25. 

3. sin^l=ii; tanJ.=£i; cotA—{%', secJ.= £i; cscJ.=f£. 

4. sin^.=0.96; tan ^. = 3.4286; cotJ.=0.2917; sec^i=3.5714; 
csc^4 = 1.0417. 

5. sin^4=0.8;cos^i=0.6; cot J.=0.75; secJ. = 1.6667; csc^4 = 1.25. 

6. sinJ. = iV2; cos^L = iV2; tan^. = l; secJ. = V2; cscJ.= V2. 

7. sin./i = 0.90; cos-4 = 0.45; tanJ. = 2; sec^i = 2.22; cbcJ. = 1.11, 

8. sin^. = iV3; cosJ. = i; tanJ.= V3; cotJ. = |V3; cscJ. = §V3 

9. sinJ. = iV2; cosJ. = ^V2; tanJ. = l; cot^. = l; secJ. = V2. 

10. cos^L = Vl — m 2 ; tan^. = — Vl - m 2 ; cot J. = — Vl — m 2 ; 

1 — ra 2 m 

secJ.= — ; csc-4 = — • 

m 

11. cos^l= -; tan^l = -; cot A = ; 

1 - m 2 2 m 

1 +m 2 



Vl - m 2 
1 


Vl — m 2 
1 - m 2 


1 + m 2 ' 
1 + w 2 


1-m 2 ' 
m 2 — n 2 


m 2 + n 2 ' 
m 2 + n 2 



sec -4 = : esc A = 



2m 



. m 2 — n 2 . 2 mn 

12. sm^L= ; ta,nA = ; cot^l = -; 

2 mn m 2 — n 2 

. m 2 + n 2 

secA=— ; csc^l = 

2 mn m 2 — n 2 

13. sin=|V2; cos=iV2; cot = 1 ; sec=V2; esc = V2. 

14. cos = \ V3 ; tan = i V3 ; cot = V3 ; sec = f V3 ; esc = 2. 

15. sin=|V3; cos = |; tan=V3; cot = ^V3; sec = 2. 

16. sin = |- V2-V3; cos = \ V2 + V3 ; cot = 2 + v^ ; 
sec = 2 (2 - V3) V2 + V3 ; esc = 2 (2 + V3) V2-V3. 



ANSWERS 



17. sin = | V2- V2 ; cos = £ V2+V2 ; tan = V2 - 1 ; 
sec = (2 - V2) V2+V2 ; esc = (2 + V2) V2 - V<2. 

18. cos = 1 ; tan = ; cot = 00 ; sec = 1 ; esc = 00. 

19. cos == ; tan = 00 ; cot = ; sec = 00 ; esc = 1. 

20. sin = 1 ; cos = ; cot = ; sec = oo ; esc = 1. 

n r-JT sin A 4 . Vl - sin 2 A 

21. cos A — vl - sin 2 ^4; tan A - - ; cot A — — - 

Vl -sin 2 ^4 

1 

sec A = ; esc A 

Vl - sin 2 A 

29 Bin /I - r-x 5~T , >. Vl - COS 2 A . COS A 

zz. sm^i — Vl - cos 2 .4 ; tan A = — ; cat A = 

Vl - cos 2 A 

sec A = ; esc -4 

cos -4 

. . tan A 

sin A = ; cos .4 = ; cot A 



sec J. = Vl + tan 2 .4 ; esc A 



:. sin A = — — : cos .A = — — ; tan A — 



VI 

1 

sin. 


j 

— sin 2 A 
A 


vT 


— cos 2 A 


cos A 
1 


vT 


— cos' 2 A 

1 


VT 


+ tan 2 J. 


Vl + tan 2 A 


tan A 
cot A 


vT 


+ cot 2 A 



Vl-j-tan 2 ^' Vl + tan 2 A ' tan J.' 



Vl + cot 2 ^l' Vl + cot 2 .4 ' cot ^ 

Vl + cot 2 ^4 

sec A = ; csc^4 = Vl -f cot 2 .4. 

cot A 

sin A = i V5 ; cos ^4 = § V5. 27. sin J. = j\ ; cos J. = £f 

• . , /TT /T7 1 -3 cos 2 ^4 +3 cos* ^4 

sin A = Wl5 ; tan A = Vl5. 28. 

cos 2 A — cos 4 A 



Exercise VII. Page 18 

1. x = 45°. 6. x = 45°. 11. x = 30°. 16. * = 45°. 

2. x = 30°. 7. x = 45°. 12. x = 45°. 17. x = 60°. 

3. x = 0°, or 60°. 8. x = 45°. 13. x = 0°, or 60°. 

4. x = 45°. 9. x = 60°. 14. x = 30°. 

5. x = 60°. 10. x = 60°. 15. x = 30°, or 45°. 



PLANE TRIGONOMETRY 



1. c 



cos A 
5. A = 90° -B; 



6. A = 90° -B; 

7. A = 90°-B; 

8. cosA = -- 



Exercise VIII. Page 24 



2. c 



a 



sin .4 

a = c cos J5 ; 

a = 6 cot 2? ; 
b = a tan B ; 
5 = 90° - A : 



3. 6 = ccosJ.. 



4. c = 



a 

sin^. 



& = c sin J3. 
6 



c = 



sin 5 

a 
cos 5 



a = V(c + b) (c - 6). 



Exercise 



31. c = 7.8112 

32. 6 = 69.997 

33. a = 1.1886 

34. 6 = 21.249 

35. a = 6.6882 

36. a = 63.859 

37. a = 19.40; 

38. 6 = 53.719; 

39. a = 12.981; 

40. a = 0.58046 

41. F=ic 2 sinAcosA. 

42. F= $ a 2 cot A. 

45. 6 = 11.6; c = 15.315; A 

46. a = 7.2; c = 8.7658 ; A 

47. a = 3.6474; 6 = 6.58; c 

48. a =10.283; 6=19.449; A 

49. 19° 28' 17" and 70° 31' 43". 

50. 3 and 5.1961. 

90° 

51. a = ccos 



A = 39° 48' 
A = 30' 12" 
A = 43° 20' 
c = 22.372 
c = 13.738 
6 = 23.369 
6=18.778 
c = 71.377 
c = 15.796 
6 = 8.442 ; 



IX. Page 28 

B = 50° 12' ; 
B = 89° 29' 48' 
B = 46° 40' ; 
B = 71° 46' ; 
B = 60° 52' ; 
B = 20° 6' ; 
A = 45° 56' ; 
A = 41° 11' ; 
A = 55° 16' ; 
^1 = 3° 56' ; 
43. F = $b 2 tan A. 



F=lb. 
F= 21.525. 
F= 0.74876. 
2^=74.372. 
F= 40.129. 
F= 746.15. 
F= 182.15. 
F= 1262.4. 
2^=58.416. 
2^= 2.4501. 



6 = c sin 



n + 1 

90° 

t + 1 



44. F=laVc 2 -a?. 
40° 45' 48" ; B = 49° 14' 12". 
55° 13' 20" ; B = 34° 46' 40". 
7.5233 ; J5 = 61°. 

27° 52' ; B = 62° 8'. 

52. 36° 52' 12" and 53° 7' 48". 

53. 212.1 feet. 

54. 732.22 feet. 

55. 3270 feet. 

56. 37.3 feet. 



ANSWERS 



57. 1°25'56". 58. 59° 44' 35" 

60. 7.0712 miles in each direction. 

61. 20.88 feet. 63. 685.9 feet. 

62. 56.65 feet. 64. 136.6 feet. 



59. 95.34 feet. 

65. 140 feet. 

66. 84.74 feet. 



Exercise X. Page 33 



1. 


C = 2(90°-^L); c 


= 2 a cos A ; ft = a sin 


^4. 




2. 


A = \ (180° -C); c 


= 2a cos ^4 ; h = asm A. 




3. 


C = 2(90° -A); a 


c * 


A. 




2 cos .4. 




4. 


A=\ (180° - C) ; 


c 


ft = a sin ^4. 




2 cos -4 ' 




5. 


C = 2 (90° - A) ; 


ft 
a = ^— -; 
sin A 


c = 2 a cos J.. 




6. 


A = i (180° - C) ; 


h 

a = - — -; 

sin A 


c = 2 a cos ^4. 




7. 


smA = ~; 
a 


C = 2 (90° - A) ; 


c = 2 acos^i. 




8. 


tan A = — ; 
c 


C = 2 (90° - .4) ; 


ft 

a = 

sin A 




9. 


A = 67° 22' 50" ; 


C = 45° 14' 20" ; 


A = 13.2. 




10. 


c = 0.21943; 


ft = 0.27384; 


F = 0.03004. 




11. 


a = 2.055 ; 


ft = 1.6852 ; 


F= 1.9819. 




12. 


a = 7.706; 


c = 3.6676 ; 


F = 13.725. 




13. 


A = 79° 36' 30"; 


C = 20° 47' ; 


c = 2.4206. 




14. 


A = 77° 19' 11"; 


C = 25° 21' 38" ; 


a = 20.5. 




15. 


^1 = 25° 27' 47"; 


C = 129° 4' 26" ; 


a = 81.41; 


ft = 35. 


16. 


^4 = 81° 12' 9" ; 


C = 17° 35' 42" ; 


a = 17 ; 

22. 0.76536. 


c = 5.2 


17. 


F = |cV4a 2 -c 2 . 




18. 


F = a 2 sin £ C cos J C. 


23. 94° 20'. 




19. 


F = a 2 sin ^1 cos J.. 




24. 2.7261. 




20. 


F = ft 2 tan £- C. 




25. 38° 56' 33' 




21. 


28.284 feet; 4525.44 


square feet. 


26. 37.699. 





PLANE TRIGONOMETRY 



Exercise XI. Page 35 



1. 


r= 1.618; 


ft = 


= 1.5388; 


Fz= 


7.694. 






2. 


h = 0.9848 


2>.= 


= 6.2514; 


F = 


3.0782. 






3. 


h = 19.754 


c = 


= 6.257; 


F = 


: 1236. 






4. 


r = 1.0824 


C : 


= 0.82842 ; 


F = 


: 3.3137. 






5. 


r = 2.5933 


h-- 


= 2.4882 ; 


c = 


: 1.4615. 






6. 


r = 1.5994 


h = 


= 1.441 ; 


P = 


: 9.716. 






7. 


0.61803. 




11. 0.2238. 






16. 


11.636 


8. 


0.64984. 




12. 0.310. 






17. 


99.640 


9. 


0.51764. 




13. 0.82842. 




18. 


1.0235. 


0. 


b- C 


o 


14. 94.63. 

15. 414.97. 






19. 


0.635. 




o 9° 





Exercise XII. Page 45 

5. Two angles ; one in Quadrant I, one in Quadrant II. 

6. Four values ; two in Quadrant I, two in Quadrant IV. 

7. x may have two values in the first case, and one value in each of the 

other cases. 

8. If cos x = — |, x is between 90° and 270° ; if cot x = 4, x is between 

0° and 90° or between 180° and 270° ; if sec x = 80, x is between 0° 
and 90° or between 270° and 360° ; if esc x = - 3, x is between 
180° and 360°. 

9. In Quadrant III ; in Quadrant II; in Quadrant III. 

10. 40 angles ; 20 positive and 20 negative. 

11. +, when x is known to be in Quadrant I or IV ; — , when 

to be in Quadrant II or III. 

12. since = + !V2; tanx = -l 



13. 



14. 



esc x = + v2. 
sin x = — \ V3 ; 
esc x = — f V3. 
sin x = — f V3 ; 



cos x = — \ ; 



cosce 



CSC£=- T 7 2 V8. 

15. sin x = ± T V VlO ; cos x = T T 3 o 
esc x = ± VlO. 



cot x = — 1 ; sec x 

cot x = + 1 V3 ; sec x 

tan x = — 4 V3 ; cot x 

VlO ; tan x = — £ ; sec x 



x is known 
= -V2; 
= -2; 

= -tVV3; 

=TlVio ; 



ANSWERS 9 

16. The cosine, the tangent, the cotangent, and the secant are negative 
when the angle is obtuse. 

18. sin 90° =1, cos 90° =0, cot 90° =0, sec 90° = oo, 
esc <;0° = 1 ; 

sin 180° = 0, tan 180° = 0, cot 180° = oo, sec 180° = - 1, 

esc 180° = oo ; 

sin 270° = - 1, cos 270° = 0, tan 270° = oo, sec 270° = oo, 

esc 270° = - 1 ; 

sin 360° = 0, cos 360° = 1, tan 360° = 0, cot 3G0° = x, 

sec 360°= 1. 

19. sin 450° = 1 ; tan 540° = ; cos 630° = ; cot 720° = oo ; 
sin 810° = 1 : esc 900° = oo. 



20. 0. 



21. 0. 



22. 0. 



23. a--b 2 + Aab. 







Exercise XIII. Page 


51 


1. 


- cos 20° 


8. 


— sin 


24°. 




15. -cos 15° 33'. 


2. 


sin 8°. 


9. 


cos 


1°. 




16. cot 0° 45'. 


3. 


- sin 10°. 


10. 


- cot 30°. 




17. -cot 40° 43'. 


4. 


- cot 35°. 


11. 


tan 6°. 




18. esc 29° 45'. 


5. 


- tan 1°. 


12. 


— esc 


26°. 




19. sec 2° 25'. 


6. 


- esc 20°. 


13- 


— sec 


1°. 






7. 


esc 23°. 


14. 


sin 


16° 11 


/ 




20. 


sin (- 75°)=- 


- cos 15° ; 




23. 


sin i 


- 345°) = sin 15° ; 




cos (— 75°) = 


sin 15° ; 






cos ( 


- 345°) = cos 15° ; 




tan(- 75°) =- 


- cot 15° ; 






tani 


-345°) = tan 15°; 




cot (-75°) = - 


■ tan 15°. 






cot( 


- 345°) = cot 15°. 


21. 


sin (-127°) = 


- cos 37° ; 




24. 


sin i 


- 52° 37') =- cos 37° 23'; 




cos(- 127°) = 


- sin 37° ; 






COS 1 


- 52° 37') = sin 37° 23'; 




tan(- 127°) = 


cot 37° ; 






tani 


-52° 37') = -cot 37° 23'; 




cot (-127°) = 


tan 37°. 






cot 1 


-52° 370 =- tan 37° 23'. 


22. 


sin (- 200°) = 


sin 20° ; 




25. 


sin i 


-190° 54') = - sin 16° 54'; 




cos (-200°) = 


-cos 20°; 






COS i 


-196° 54')= -cos 16° 54'; 




tan (-200°) = 


- tan 20° ; 






tani 


-196° 54')= -tan 16° 54'; 




cot (-200°) = 


- cot 20°. 






COt I 


-196° 540= -cot 16° 54'. 



10 PLANE TRIGONOMETRY 

26. sinl20°=+iV3; cosl20°=-^; tanl20°=-V3; cotl20°=-iV3. 

27. sin 135°=+ 1 V2; cos 135°= -| Vjj ; tan 135°= -1; cot 135°= -1. 

28. sin 150°= i; cos 150°= -| VI; tan 150°= -^V3; cotl50°=-V3. 

29. sin210°=-|; cos210°= -\ V§; tan210°=+i V3; cot210 D = + V3. 

30. sin225°=-^V2; cos225°=-| V2; tan225°=l; cot225°=l. 

31. sin240°=-iV3; cos240°=-£; tan240°= + V3; cot240°=+i Vs. 

32. sin 300°= -^V3; cos300°=i; tan300°=-V3; cot300°= -iV3. 

33. sin (- 30°) = - \ L cos(- 30°) = + \ V3 ; tan (- 30°) =- \ Vg; 
cot(-30°) = -V3. 

34. sin (- 225°) = + | V2 ; cos (- 225°) = - | V£ ; tan (- 225°) = - 1 ; 
cot(-225°)=-l. 

35. cos x = — \ V2 ; tan x = 1 ; cot x = 1 ; x = 225°. 

36. sinx = |; cosx = - \ V3; tanx=-i V3; x = 150°. 

37 . sin 3540° = - 1 V3 ; cos 3540° = a ; tan 3540° = - V3 ; 
cot 3540° = - i Vs. ' 

38. 210° and 330° ; 120° and 300°. 

39. 135°, 225°, and - 225°; 150° and - 30°. 

40. 30°, 150°, 390°, and 510°. 

41 . sin 168° ; cos 334° ; tan 225° ; cot 252° ; 
sin 349°; cos 240°; tan 64°; cot 177°. 

42. 0.8480. 43. -1.9522. 44. (a -b) sin x. 

45. msinxcosx. 48. 0. 

46. (a — b) cot x — (a + b) tan x. 49. cos x sin y — sin x cos y. 

47. a 2 + 6 2 + 2a6cosx. 50. tan x. 

51. Positive between x = 0° and x = 135°, and between x = 315° and 

x = 360° ; negative between x = 135° and x - 315°. 

52. Positive between x — 45° and x — 225° ; negative between x — 0° and 

x - 45°, and between x = 225° and x = 360°. 

53. sin (x — 90°) = — cos x ; cos (x — 90°) = sin x ; 
tan (x - 90°) = - cot x ; cot (x - 90°) = - tan x. 

54. sin (x - 180°) = - sin x ; cos (x - 180°) = - cos x ; 
tan (x - 180°) = tan x ; cot (x - 180°) = cot x. 



ANSWERS 



11 



Exercise XIV. Page 60 

sin (x + y) = £§ ; cos (x + y) = f §. 



cosy 



cosy 
coty 



4. 

5. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 
14. 

15. 



sin ( 90° - y) 

sin ( 90° + y) 
tan ( 90° + y) 

sin (180° — y) = sin y 
tan (180° -y)=- tany 

sin (180° + y) = - sin y 
tan (180° 4- y) = tany 
sin (270° — y) = — cosy 
tan (270° - y) - cot y 

sin (270° + y) = — cosy 
tan(270° + y)=-coty 

sin (360° - y) = - sin y 
tan (360° - y) = - tan y 

sin (360° + y) = sin y 
tan (360° + y) = tan y 

sin (x — 90°) = — cosx 
tan(x- 90°) = -cotx 

sin (x — 180°) = — sin x 
tan (x - 180°) = tan x 

sin (x — 270°) = cos x 



cos ( 90° — y)= sin y. 

cos ( 90° + y) = — sin y ; 
cot ( 90° + y) - - tan y. 

cos (180° — y) = — cos y ; 
cot (180° - y) = - cot y. 

cos (180° + y) = — cos y ; 
cot (180° + y) = coty. 

cos (270° — y) = — sin y ; 
cot (270° - y) = tan y. 

cos (270° + y) = sin y ; 
cot (270° + y) = - tan y. 

cos (360° - y) = cos y ; 
cot (360° - y) - - cot y. 

cos (360° + y) = cos y ; 
cot (360° + y) = cot y. 

cos (x — 90°) = sin x ; 
cot (x - 90°) = - tan x. 

cos (x — 180°) = — cos x ; 
cot (x - 180°) = cot x. 

cos (x — 270°) = — sin x ; 
cot (x - 270°) = - tan x. 



tan (x - 270°) = - cot x ; 

sin (— y) = — sin y ; cos (— y) = cos y ; 
tan (— y) = — tany ; cot (— y) = — cot y. 

sin (45° — y)= iV2(cosy — siny) ; cos(45°— y) = | V2 (cos y + sin y) : 



tan (45° - y) = 



1 — tan y 



cot (45°- y) 



cot y + 1 



1 + tan y cot y — 1 

sin (45° + y)= 1V2 (cos y + siny) ; cos(45°+y) = i V2 (cosy— sin 



tan (45° + y) 



1 + tan y 



cot (45°+ y)- 



coty 



16. sin 



1 — tan y cot y + 1 

4- y) = 1 (cos y -f V3 sin y) ; cos (30°+. y) = |( V3 cos y — sin 



tan (30° 4- y) = 



i V3 4- tany 
L - i V3 tan y 



cot(30°+y): 



V3coty- 1 
cot y 4- v^ 



12 PLANE TRIGONOMETRY 

17. sin (60°-y)=^(V3cosy-siny); cos (60°- y) = |(cosy+ Visiny); 

tan(60°- y) = V5 - tany ; cat«W- y ) = * V5eot » + .. 1 . 

1 + v3 tan y cot y - i V.3 

18. 3 sin x - 4 sin 3 x. 19. 4 cos 3 x - 3 cos x. 20. 0. 21. i V3. 



22. sin ix = V 1 ^-^ = °- 10051 5 cos ^x = \ * + ° A V ° = 0.99493. 

23. cos2x = - i tan2x =— V3. 



24. sin 22 1° = \ V2 -V2 = 0.3827 ; cos 22|° = 1 V2+V2 = 0.9239 ; 
tan 22 \° = V2 - 1 = 0.4142 ; cot 22^° = V2 + 1 = 2.4142. 

25. sin 15° = \ V2 - V3 = 0.2588 ; cos 15° = \ V2 + V3 = 0.9659 ; 
tan 15° = 2 - Vs = 0.2679 ; cot 15° = 2 + V3 = 3.7321. 

34. sin A + sin B + sin C = sin ^4 -f- sin B + sin [180° - (A + B)] 

= sin A + sin B + sin (J. + B) 
By [20] and [12], 

= 2 sin \{A + 5) cos \(A - B) + 2 sin £(4 + 5)cos|(^ + J5) 

= 2 sin \ (A + B) [cos 1 (A - B) + cos \{A + J5)] 
By [22], 

= 2 sin 1 (A + B) (2 cos £ J. cos 1 B) 

= 4 sin 1 (A + #) cos | A cos i 5. 
But cos i C = cos [90° -\(A + B)] = sin £ (-4 + £). 





.-. sin A + 


sin 


B + sin C = 4 cos 


M 


cos i B cos 1 C. 


35. 


Proof similar to that for 34. 






38 


2 




42. tan 2 x. 
13 cos(x-y) 




46 cos(x + y) 




sin 2 x 
2 cot 2 x. 


sin x sin y 


39. 


cos x cos y 


47. tan x tan y. 


40. 


cos (x — y) 
sin x cos y 




cos (x + y) 
cos x cos y 






41. 


cos (x + y) 
sin x cos y 




15 cos(x-y) 






sin x sin y 





Exercise XV. Page 63 

1. sin-H V3 = 60° + 2 ntf or 120° + 2n7T; 
tan-H V3 = 30° + 2 W7T or 210° + 2 nit ; 
vers- 1 i = 60° + 2 wtt or 300° + 2 nn ; 



ANSWERS 13 

cos-^-i V2) = 135° + 2 nit or 225° + 2 nit ; 

esc- 1 V2 = 45° + 2 nit or 135° + 2 nit ; 

tan- 1 oo = 90° + 2 n7T or 270° + 2nit; 

sec- 1 2 = 60° + 2 nit or 300° + 2 war ; 

cos- 1 (- f Vg) = 150° + 2 nnr or 210° + 2 titt. 

4. ±V2. 10. ± T 5 3. 12. ±iV2. 

8. 0°, 90°, 180°. 11. ± &. 13. x = or ± \ Vz. 

Exercise XVI. Page 67 

1. If, for instance, C = 90°, [25] becomes - = sin A. 

3. a 2 = 6 2 + c 2 ; a 2 = 6 2 + c 2 - 2 6c ; a 2 = 6 2 + c 2 + 2 be ; a right triangle ; 
a straight line ; a straight line. 

4. b = a cos C + c cos A ; a = 6 cos C + c cos 2? ; c = b cos J.. 

6. 90°. 

7. (i) = tan (A — 45°); a right triangle. 

a + b 

(ii) a + 6 = (a — 6) (2 + V3) ; an isosceles triangle with the angles 30°, 
30°, 120°. 

Exercise XVII. Page 69 

9. 300 yards. 15. a = 5 ; c = 9.6593. 

10. ^45 = 59.564 miles; 16. a = 1; 6 = 8.573. 

AC = 54.285 miles. 17. sides, 600 feet and 1039.2 feet ; 

11. 4.6064 miles; 4.4494 miles; altitude, 519.6 feet. 
3.7733 miles. 18. 855:1607. 

12. 4.1501 and 8.67. 19. 5.433 and 6 . 857# 

13. 6.1433 miles and 8.7918 miles. 20. 15.588. 

14. 8 and 5.4723. 

Exercise XVIII. Page 74 

1. Two; one; no solution ; one; two; no solution ; one. 
11. 420. 12. 124.617. 



14 



PLANE TRIGONOMETRY 



Exercise XIX. Page 78 

11. 6. 15. 25. 18. 10.266 miles. 

12. 10.392. 16. 3800 yards. 19. 5.0032 and 2.3385. 

14. 8.9212. 17. 729.67 yards. 20. 26° 0' 10" and 14° 5' 50' 

21. 430.85 yards. 



Exercise XX. Page 83 

11. A = 36° 52' 12" ; B = 53° 7' 48" ; C = 90°. 

12. A = B = 33° 33' 27" ; C= 112° 53' 6". 

13. A = B=C = 60°. 

14. A = 28° 57' 18" ; B = 46° 34' 6 

15. A = 45° ; £ = 120° ; = 15°. 

16. A = 45° ; 5 = 60° ; C = 75°. 

17. 4° 23' 2" W. of N., or W. of S. 

18. 60°. 
20. 0.88877. 



C = 104° 28' 36". 

21. 54.516 miles. 

22. 84° 14' 34". 

23. 54° 48' 54". 

24. 105°; 15°; 60°. 

25. 12.434 inches. 



Exercise XXI. Page 87 

1. 4,333,600. 6. 26,208. 11. 0.19975. 

2. 365.68. 7. 15,540. 

3. 13,260. 8. 29,450 or 6982.8. 

4. 8160. 9. 17.3206. 

5. 240. 10. 10.392. 



12. F=absmA. 

13. F=£(a 2 -&2)tan^ 

14. 2,421,000. 

15. 30°; 30°; 120°. 



Exercise XXII. 

1. 21.166 miles; 24.966 miles. 

2. 6.3399 miles. 

3. 119.29 feet. 



Page 88 

4. 30°. 

5. 20 feet. 

6. 2.6247 or 21.4587. 



1. 106.70 feet; 
142.86 feet. 

2. 1023.9 feet. 



Exercise XXIII. Page 90 

3. 37° 34' 5". 6. 2922.4 miles. 

4. 238,410 miles. 7. 60°. 

5. 861,860 miles. 8. 3.2068. 









ANSWERS 




1 


9. 


6.6031. 


31. 


13.657 miles per 


47. 


6.3397 miles. 


10. 


199.56 feet. 




hour. 


48. 


210.44 feet. 


11. 


43.107 feet. 


33. 


56. 564 feet. 


50. 


757.50 feet. 


12. 


45 feet. 


34. 


51.595 feet. 


51. 


520.01 yards. 


13. 


26° 34'. 


35. 


101.892 feet. 


52. 


1366.4 feet. 


14. 


78.367 feet. 


37. 


N. 76°56 , E.; 


53. 


658.36 pounds; 


15. 


75 feet. 




13.938 miles per 




22° 23' 47" with 


16. 


1.4446 miles. 




hour. 




first force. 


17. 


7912.8 miles. 


38. 


442.11 yards. 


54. 


88.326 pounds; 


18. 


56.649 feet. 


39. 


255.78 feet. 




45° 37' 16" with 


19. 


69.282 feet. 


40. 


3121.1 feet; 




known force. 


■20. 


260.21 feet; 




3633.5 feet. 


57. 


536.28; 500.16. 




3690.3 feet. 


41. 


529.49 feet. 


58. 


345.48 feet. 


21. 


1.3438 miles. 


42. 


41.411 feet. 


59. 


345.46 yards. 


22. 


235.81 yards. 


43. 


234.51 feet. 


60. 


61.23 feet. 


26. 


8.0076 inches. 


44. 


25.433 miles. 


62. 


307.77 yards. 


29. 


460.46 feet. 


45. 


294.69 feet. 


63. 


19.8; 35.7; 44.5 


30. 


88.936 feet, 
cos A = 


46. 


12,492.6 feet. 


64. 


45°, 135°, 225°, 
or 315°. 


65. 


Vw2 + 


4(n + l) 



15 



67. 
68. 
69. 

71. 

72. 
73. 
74. 
76. 
77. 
78. 
79. 
80. 
81. 
82. 



sin A 



Vm 2 -n 2 _ n ll^ 
; cos B = — \ I— 
l-n a ' ro \1- 



-m 2 



60°, 120°, 240°, or 300°. 

0°, 60°, 180°, or 300°. 

0°, 30°, 150°, 180°, 210°, 330 c 



a 180° _ 

- cot : R 

2 n 



a 180 c 

-CSC 

2 n 



F = $ be sin A. 

F = £c 2 sin A sin B esc (A + B). 
F = Vs(s-"a)(s-6)(s-c). 
199 acres 8 square chains. 
210 acres 9.1 square chains. 
12 acres 9.78 square chains. 

3 acres 0.392 square chains. 
12 acres 3.45 square chains. 

4 acres 6.634 square chains. 
14 acres 5.54 square chains. 



83. 61 acres 4.97 square chains. 

84. 4 acres 6.633 square chains. 

85. 13.93 chains; 23.21 chains; 
32.50 chains. 

86. 9 acres 0.055 square chains. 

88. 876.34. 

89. 1229.5. 

91. 1075.3. 

92. 2660.4. 

93. 16,281. 

94. 435.76 square feet. 

95. 49,088 square feet. 

96. 749.95 square feet. 

97. 422.38 square feet. 



16 PLANE TRIGONOMETRY 

98. 1834.95 square 108. 6086.4 feet. 111. 228.98 miles; 
feet. 109. 5° 25' 6" S. ; 11° 39' 6" S. 

99. 26.88. 457.49 miles. 112. S. 56° 7' 32" E. ; 
102. 6. 110. 460.79 miles; 202.58 miles. 
107. 6. 383.13 miles. 

113. N. 17° 25' 22" W. ; 119. 33° 18' 22" N. ; 36° 23' 53" W. 
37° 46' 13" N. 120. N. 28°47 , 26"E.; 1292.8 miles. 

114. 244.35 miles ; S. 56° 10' 49"E. 121. S. 50° 39' 44" W. ; 

115. 359.87 miles. 250.84 miles ; 20° 9' 30" W. 

117. Long. 68° 54' 39" W. 122. 38° 20' 34" N. ; 55° 12' 4" W. 

118. 103.57 miles. 123. 171.14 miles; 32° 43' 38" W. 

124. N. 36° 52' 12" W. ; 36° 7' 37" W. 

125. 173.18 miles ; 51° 16' 16" S. ; 34° 12' 43" E. 

126. S. 50° 57' 48" E. ; 47° 14' 35" N. ; 20° 48' 37" W. 

127. N. 53° 20' 21" E., 16° 6' 57" W. ; or N. 53° 20' 21" W., 25° 53' 3" W. 

128. N. 47°42'33" E., 19°27'22"N., 121°50'34"E. ; or N. 47°42'33" W., 

19° 27' 22" N., 116° 9' 26". E. ; or S. 47° 42' 33" E., 14° 32' 38" N., 
121°48'20"E. ; or S. 47° 42' 33" W., 14° 32' 38" N., 116° 11' 40" E. 

129. 359.82 miles; 359.73 miles; 359.50 miles. 

130. 35° 49' 10" S., 22° 2' 44" W. ; N. 61° 42' W. ; 183.16 miles. 

131. 42° 15' 29" N., 69° 5' 11" W.; N. 72° 32' 40" E.; 44.939 miles. 

132. 32° 53' 34" S., 13° 1' 53" E.; N. 72° 3' 43" W.; 287.16 miles. 

Exercise XXIV. Page 107 

(The solutions here given are for angles less than 360°.) 

79. sin£x = ±£ V5; C os£cc = ±f VE. 81. ± £ V3. 

80. ± V5 - 2. 82. ± I or ± f . 

83. ±2V2, ±^(9Vs + 8V2),or ± J ¥ (9V3-8V2). 

84. ±h 85. i(V5-l); i(V5 + l). 

86. (at + b*)i. 91. 4- a + 1 . 96. tan-i-^-. 

h V2a + 1 l-2x 2 

87. (i^\ (lT2mV 92. 4. 97. 2. 

88. ± \ V2 or ± $ V3. 93. tan (x + y). 98. - tan 4 x + cot 4 x. 

89. f. 94. 99. x = $7t or %7t. 

sin y 

90. f or - |. 95. - tan x. 100. x = 90° or 270°. 



ANSWERS 17 

101. x = 21° 28' or 158° 32'. 

102. x = 0°or90°. 

103. x = 30°, 150°, 199° 28', or 340° 32'. 

104. x = 51° 19', 180°, or 308° 41'. 

105. x = 0°, 120°, 180°, or 240°. 

106. x = 45°, 161° 34', 225°, or 341° 34'. 

107. = 60°, 120°, 240°, or 300°. 

108. 6 = 26° 34' or 206° 34'. 110. x = 45° or 135°. 

109. x = 30° or 150°. 111. x = 30°, 150°, or 270°. 

112. x = 35° 16", 144° 44', 215° 16', or 324° 44'. 

113. x = 75° 58' or 255° 58'. 

114. 6 - 60°, 180°, or 300°. 116. x = 30°, 150°, 210°, or 330 c . 

115. 6 = 90° or 143° 8'. 117. x = 30°, 160°, or 270°. 

118. x = 26° 34', 90°, 206° 34', or 270°. 

119. x = 45°, 136°, 225°, or 315°. 

120. x = 45°, 135°, 225°, or 315°. 

121. x = 15°, 75°, 135°, 195°, 255°, or 315° 

122. z = 45°, 135°, 225°, or 315°. 

123. x = 0°, 60°, 120% 180°, 240°, or 300°. 

124. x = 27° 58', 135°, 242° 2', or 315°. 

125. x = 0°, 45°, 180°, or 225°. 

126. x = 32° 46', 147° 14', 212° 46', or 327° 14'. 

127. x = 0°, 45°, 90°, 180°, 225°, or 270°. 

128. x = 0°, 65° 42', 180°, or 204° 18'. 

129. x = 0°, 90°, 120°, 240°, or 270°. 

130. x = 0°, 36°, 72°, 108°, 144°, 180°, 216°, 252°, 288°, or 324°. 

131. x = 30°, 150°, 210°, or 330° 

132. x = 60° or 240°. 

133. x = 54° 44', 125° 16', 234° 44', or 305° 16'. 

134. x = 105° or 345°. 
a 2 -l 



135. x = tan- 1 

136. x = cos 



la 

1 -g = bVa 2 + 8a + 8 



4 



18 PLANE TRIGONOMETRY 

137. x = 135°, 315°, or isin-i (i _ a y 

138. x = 30°, 60°, 120°, 150°, 210°, 240°, 300°, or 330°. 

139. x = 60°, 90°, 120°, 240°, 270°, or 300°. 

140. x = 60°, 90°, 120°, 240°, 270°, or 300°. 

141. x = 120°. 142. x = 14° 29', 30°, 150°, or 165° 31'. 

143. x = 60°, 90°, 270°, or 300°. 

144. x = 0°, 20°, 100°, 140°, 180°, 220°, 260°, or 340°. 

145. x = 45°, 90°, 135°, 225°, 270°, or 315°. 

146. x = 30°, d0°, 90°, 120°, 150°, 210°, 240°, 270°, 300°, or 330°. 

147. x = 0°, 45°, 90°, 180°, 225°, or 270°. 

148. x = 30°, 60°, 120°, 150°, 210°, 240°, 300°, or 330°. 

149. x = 30°, 90°, 150°, 210°, 270°, or 330°. 

150. x = 0°, 45°, 180°, or 225°. 

151. x = 45°, 60°, 120°, 135°, 225°, 240°, 300°, or 315°. 

152. x = 0°, 45°, 135°, 225°, or 315 Q . 

153. x = 30°, 90°, 150°, 210°, 270°, or 330°. 

154. x = 8° or 168°. 

155. x = 40° 12', 139° 48', 220° 12', or 319° 48'. 

156. x = 30° or 330°. 

157. x = 60°, 120°, 240°, or 300°. 

158. x = 30°, 60°, 120°, 150°, 210°, 240°, 800°, or 330°. 

159. x = 53° 8', 126° 52', 233° 8', or 306° 52'. 

160. x = 30°. 161. x = 22° 37' or 143° 8'. 

162. x = 0°, 20°, 40°, 60°, 80°, 100°, 120°, 140°, 160°, 180°, 200°, 220°, 240°, 

260°, 280°, 300°, 320°, or 340°. 

163. x = 22° 30', 45°, 67° 30', 90 6 , 112° 30', 135°, 157° 30', 202° 30', 225°, 

247° 30', 270°, 292° 30', 315°, or 337° 30'. 

164. x = 45° or 225°. 169. x = 1. 

165. x = ± 1 or ±i V21. 170. x = or £ 1. 

166. x = i V3 or - 1 V§. 171. x = ± £ V2. 

167. s = £V3. 172. x = }V3. 

168. x = h 173. = 120° or 240°. 

174. x = 60°, 120°, 240°, or 300°. 

175. x = 0°, 45°, 135°, 225°, or 315°. 



ANSWERS 



19 



176. 


x = 


177. 


X = 


178. 


6 -- 


179. 


X - 


180. 


X = 


181. 


e = 


182. 


X - 


183. 


X - 


184. 


X = 


185. 


-. 




or 


186. 


x = 


189. 


X z 



= 0°, 180°, 220° 39', or 319° 21'. 
= 0°, 60°, 120°, 180°, 240°, or 300°. 
3 18°, 90°, 162°, 234°, 270°, or 306°. 
= 0°, 30°, 90°, 150°, 180°, 210°, 270°, or 330°. 
r 0°, 90°, 120°, 180°, 240°, or 270°. 
= 0°, 74° 5', 127° 25', 180°, 232° 35', or 285° 55'. 
= 0°, 90°, 180°, or 270°. 
= 0°, 45°, 90°, 180°, 225°, or 270°. 
= 0°, 45°, 120°, 135°, 180°, 225°, 240°, or 315°. 
= 10° 12', 34° 48', 190° 12', 187. = 29° 19', lbo° 41', 209° 19', 
214° 48'. or 285° 41'. 

= 90° or 270°. 188. x = 1. 

b sin 8 — a cos 8 a cos a — b sin a 

' y = 



sin (8 — a) 



sin (8 — a) 



190. 
191. 
192. 



193. 

194. 
196. 
197. 
198. 

199. 



tan- 1 - + cos- 1 \ Va 2 + b 2 ; y = tan- 1 - - cos- 1 1 Va 2 + b 2 . 
b b 



tan- 1 -; r=V a 2 + V*. 
b 



. / .acos/3 — 6sina \ 

sin ( tan -1 - \- a ) 

,a / 



= tan- 1 



asin/3-f 6cos. 
a cos 8 — b sin a 



/ ,acos/S — frsina- \ 

cos( tan- 1 + 8 ) 

,a / 



asin/3 + 6cosi 



a sin 8 + b cos a 



Va 2 -f b 2 -f c 2 ; ^ = tan~ 1 -; = tan- 1 

b Va 2 + b 2 

100 ; y = 200. 195. x = 76° 10' ; y = 15° 30'. 

225.12, = 24° 13' ; or r = - 225.12, = 204° 13'. 
151, 6 = 42° 28' ; or r = - 151, = 222° 28'. 



x 

r 

r 

r = 108 

r 



- 108, </> = 120°, = 150° ; or r = - 108, <p = 300°, = 30°. 
r vers- 1 ^- j ^y/2ry — y 2 . 



Exercise XXV. Page 121 



log 10 6 =0.77815 
log 10 4 =0.60206 
log 10 i =1.69897 
logiofi = 0.02119 



logioH =1.14613; 


logi 21 = 1.32222 


logi 12 =1.07918; 


logio5 =0.69897 


logio I = 1.39794 ; 


logi<>£ =1-89086 



20 



PLANE TRIGONOMETRY 



2. log 2 10 =3.3219; log 2 5 =2.3219; log 3 5 =1.4650 

log 7 \ = - 0.3562 ; log 5 7 f T = - 2.2620. 



3. log e 2 =0.69315; 



log e 3 



1.09862 



log* 5 



1.60945 



log e 7 =1.94593; log e 8 =2.07946; log e 9 =2.19724 

log e f =-0.40547; log e | =-0.22315; log e ff =0.25952 
loge ^ =-2.14845. 
4. x = 1.54396 ; x = 0.83048 ; x = 0.42062. 



Exercise XXVI. Page 126 

1. log e 3 =1.09861. 2. log e 5 =1.60944. 3. log c 7 =1.94591. 

4. log e 10 =2.3025850930. 

5. logi 2 =0.30103; logi e = 0.43429 ; log 10 11 = 1.04139. 



Exercise XXVII. Page 128 

1. sin r = 0.00029088820; cos 1' = 0.99999995769 ; 
tan 1' = 0.000290888212. 

2. sin 2' = 0.000581776. 3. sin 1° = 0.0175. 



6. 0°40' 9' 



Exercise XXVIII. Page 130 

1. sin 6' = 0.0017453 ; cos 6' = 0.9999985. 

2. sin 2° = 0.034902; cos 2° = 0.999392. 

3. sin 3° = 0.052339 ; cos 3° = 0.998632. 

4. sin 4° = 0.069760 ; cos 4° = 0.997564. 

5. sin 5° = 0.087160 ; cos 5° = 0.996193. 



Exercise XXIX. Page 135 



1. The 6 sixth roots of — 1 are : 

Vs-\-i . -V3 + i -V3 



h 



2 2 2 

The 6 sixth roots of + 1 are : 
1 + V3~3 _ 1+ VT"3 



1, 



. V3-i 

i, —. — 



1 -V^3 1 



Vs + i -Vs 



+ i 



ANSWERS 21 

3. cos 67£° + i sin 67£°, cos 157 4° + i sin 157£°, cos 247£° + i sin 2474°, 
cos 337^°+ i sin 337 1°. 

4. sin 4 = 4 cos 3 sin — 4 cos sin 3 ; 
cos 4 — cos 4 — 6 cos 2 6 sin 2 + sin 4 0. 



Exercise XXX. Page 137 

, x 2 5x 4 61 x 6 

5 . sec x = 1 H 1 (- — h • • • 

2 24 720 

x 2 x 4 2x 6 

6. x cot x=l 

3 45 945 

7. sin 10° = 0.173648; cos 10° = 0.984808. 

8. tan 15° = 0.267949. 



SPHERICAL TRIGONOMETRY 

Exercise XXXI. Page 142 

1. 110°; 100°; 80°. 2. 140°; 90°; 65°. 

ItT 

5 . Multiply their measures in degrees by 

6. %it feet ; 2 it feet ; -% 5 - it feet. 

Exercise XXXII. Page 146 

3. (i) Either a or b = 90° ; (ii) A = 90° and B = b • 
(iii) A = 90° and B = b ; (iv) c = 90° and b = B = 90°. 

Exercise XXXIII. Page 148 

2. Rule I. The cosine of any middle part is equal to the product of 

the cotangents of the adjacent parts. 
Rule II. The cosine of any middle part is equal to the product of 
the sines of the opposite parts. 

Exercise XXXIV. Page 153 

24. A = 175° 57' 10" ; B = 135° 42' 50" ; C = 135° 34' 7". 

25. a = 104° 53' 2" ; b = 133° 39' 48" : C = 104° 41' 39". 



22 SPHERICAL TRIGONOMETRY 

26. a = 90° ; b = B; b and B are otherwise indeterminate. 

27. a = 60°; & = 90°; £ = 90°. 

28. The triangle is impossible. 

29. b = 130° 41' 42" ; c = 71° 27' 43" ; J. = 112° 57' 2". 

30. a = 26° 3' 51"; ^4 = 35°; 5 = 65° 46'. 

31. The triangle is impossible. 

32. a = 60° 16' 17" ; b = 29° 41' 4" ; 1? = 33° 16' 54". 

33. & = 42° 10' 17" ; c = 106° 37' 37" ; A = 105° 41' 39". 

34. a = 113° 51' 5"; c = 105° 37' 54" ; £ = 50° 44' 19". 

35. a = 124° 10' 37" ; b = 107° V 22" ; c = 80° 28' 49". 

Exercise XXXV. Page 157 

1 . cos A = cot a tan \b; sin £ 5 = esc a sin £ b ; cos ft = cos a sec £ 6. 

2. sin | J. = i sec £ a. 

o ■ , a t 180 ° • i> -i 180 ° 

3. sin A J. = sec 4 a cos ; sin R = sin -J- a esc ; 

n n 

sin r = tan i a cot 

n 

4. Tetrahedron, 70° 31' 46" ; hexahedron, 90° ; octahedron, 109° 28' 14' 

dodecahedron, 116° 33' 45"; icosahedron 138° 11' 36". 



5. cot$A = Vcosa. 

Exercise XXXVI. Page 160 

1 . (i) sin a sin B = sin &, sin a sin C = sin c ; 
(ii) sin a = sin b sin ^4, sin & sin C = sin c ; 

(iii) sin a = sin c sin 4, sin 6 = sin c sin B ; 
(iv) sin 23 = sin b sin J., sin C = sin c sin J. ; 
(v) sin a = sin 6, sin c = sin a sin C = sin 6 sin C ; 
(vi) sin J5 = sin A, sin C = sin c sin J. = sin c sin J5. 

2. (i) cos a = cos b cose; (ii) cos 6 = cos a cose ; (iii) cos c = cos a cos 6 
(iv) cos a = cos b cos c, cos b = cos a cos c, cos c = cos a cos 6. 

3. (i) cos a = cos (6 — c) ; (ii) cos a = cos 6 cose; (iii) cos a = cos(6+c). 

Exercise XXXVII. Page 167 

1. (i) tan m = tan b cos J., cos a = cos b sec m cos (c — m); 
(ii) tan m = tan c cos JB, cos b = cos c sec w cos (a — m). 



ANSWERS 23 

Exercise XXXVIII. Page 169 

1. (i) cot x = tan B esc a, cos A — cos B esc x sin (C — x) ; 
(ii) cot a: = tan C esc &, cos B = cos C esc x sin (J. — x). 

Exercise XLIII. Page 180 

4. 2.2298 .R 2 . 8. 1.1891 JP. 12. 3.1416 i? 2 . 

5. 1.4956 R 2 or 0. 17085 E 2 . 9. 0.7105 R 2 . 13. 5.4206 R 2 . 

6. 0.95484 R 2 . 10. 0.09301 R 2 . 14. 2070.1 square miles. 

7. 0.024832 R 2 . 11. 2.8624 R 2 . 

Exercise XLIV. Page 197 

1. 148° 42'. 2. cos x = cos A cos 2*. 

3. Let w equal the inclination of the edge c to the plane of a and 6. 
Then it is easily shown that F = abc sin I sin w. Now, conceive 
a sphere constructed having for centre the vertex of the trihedral 
angle whose edges are a, 6, c. The spherical triangle, whose 
vertices are the points where a, 6, c meet the surface of this 
sphere, has for its sides, I, m, n; and w is equal to the perpen- 
dicular arc to the side I from the opposite vertex. Let L, M, N 
denote the angles of this triangle. Then, by [39] and [47], 
sin w = sin m sin N 

= 2 sin m sin \ N cos \ N 

2 I 

= - — Vsin s sin (s — I) sin (s — m) sin (s — n), 
sin I 

where s = $(l + m + n). 



.•. F = 2 a6c v sin s sin (s — I) sin (s — ?n) sin (s — n). 

4. (i) 9,976,500 square miles ; (ii) 13,316,560 square miles. 

5. Let m equal the longitude of the point where the ship crosses the 

equator, B her course at the equator, d the distance sailed. 
Then, tan m = sin I tan a, 

cos B = cos I sin a, 
cot d = cot I cos a. 

6. Let A; equal the arc of the parallel between the places, x the difference 

required. 
Then, x = 2 cos I sin— l (sin -J- d sec l) — d. 

x = 90°(V2- 1). 



24 SPHERICAL TRIGONOMETRY 



7. tan i (m — m') = Vsec s sec (s — d) sin (s — I) sin (s — V) ; where 2 s 
= I + V + d, and m and wi' are the longitudes of the places. 

9. 44 minutes past 12 o'clock. 10. 60°. 

11. cos£ = — tan d tan Z ; time of sunrise = 12 o'clock a.m. ; time 

t 15 

of sunset = — o'clock p.m. ; cos a = sin d sec I. For longest day 

15 
at Boston : time of sunrise, 4 hr. 26 min. 50 sec. a.m.; time of 
sunset, 7 hr. 33 min. 10 sec. p.m. Azimuth of sun at these times, 
57° 25' 15" ; length of day, 15 hr. 6 min. 20 sec. ; for shortest day, 
times of sunrise and sunset are 7 hr. 33 min. 10 sec. a.m. and 4 hr. 
26 min. 50 sec. p.m. ; azimuth of sun, 122° 34' 45" ; length of day, 
8 hr. 53 min. 40 sec. 

12. The problem is impossible when cotd<tanZ; that is, for places in 

the frigid zones. 

13. For the northern hemisphere and positive declination, 

sin h = sin I sine?, cot a = cos Z tan d ; ft = 17° 14' 35", a = 73° 51' 34" K. 

14. The farther the place from the equator, the greater the sun's altitude 

at 6 a.m. in summer. At the equator it is 0°. At the north pole 
it is equal to the sun's declination. At a given place the sun's 
altitude at 6 a.m. is a maximum on the longest day of the year, 
and then sin ft = sin I sin e (where e = 23° 27'). 

15. cos t = cot I tan d. Times of bearing due east and due west are 

12 o'clock a.m. and — o'clock p.m., respectively; 6 hr. 

15 15 * 

58 min. 10 sec. a.m. and 5 hr. 1 min. 50 sec. p.m. 

16. When the days and nights are equal, d = 0°, cos t = 0, t = 90° ; that 

is, sun is everywhere due east at 6 a.m., and due west at 6 p.m. 

Since I and d must both be less than 90°, cos t cannot be negative, 

therefore t cannot be greater than 90°. As d increases, t decreases ; 

that is, the times in question both approach noon. 
If I < d, then cos t > 1 ; therefore this case is impossible. 
If I = d, then cos t = 1, and t = 0° ; that is, the times both coincide 

with noon. The explanation of this result is, that for I = d the 

sun at noon is in the zenith, and south of the prime vertical at 

every other time. 



ANSWERS 25 

If l>d, the diurnal circle of the sun and the prime vertical of the 
place meet in two points which separate farther and farther as I 
increases. At the pole the prime vertical is indeterminate ; but 
near the pole, t = 90°, and the sun is always east at 6 a.m. 

17. sin I = sin d esc h. 18. 11° 50' 35". 

19. The bearing of the wall, reckoned from the north point of the hori- 

zon, is given by the equation cotx = cos I tan d ; whence, for the 
given case, x = 75° 12' 28". 

20. 55° 45' 6" N. 21. 63° 23' 41" N. or S. 

22. (i) cos t — — tan I cotp ; (ii) t = z ; (iii) the result is indeterminate. 

23. cot a = cos I tan d. 28. sin d = sine sin v; tan r = cose tan v. 

25. h = 65° 37' 20". 29. d = 32° 24' 12"; r = 301° 48' 17". 

26. h = 58° 25' 8"; a = 152° 28'. 30. d = 20° 48' 38". 

27. * = 45° 42'; 1 = 67° 58' 54". 31.3 hr. 59 min. 27 T 4 3 sec. p.m. 
32. cos | a = Vcos £ (I + h + p) cos £ (J + & — _p) sec J sec A. 



SURVEYING 

Exercise II. Page 230 
2. 540°. 4. N. 51° 30' E. 

Exercise III. Page 235 
2. 360°. 3. 235 ft. 3.8 in. 

Exercise V. Page 263 

1. 8 a. 64 p. 6. 13 a. 6 t l p . 11. 4 a. 35 p. 

2. 16 a. 74|f p. 7. 2 a. 58| p. 12. 4 a. 110 p. 

3. 4 a. 5& p. 8. 11 a. 157 p. 13. 6 a. 23^| p. 

4. 115^ ^ 9. 7.51925. 

5. 3 a. 78 p. 10. 13.0735. 



26 



SURVEYING 



1. 2 a. 26 p. 

2. 20 a. 12 p. 



Exercise VI. Page 272 



3. 8 a. 54 p. 

4. 3 a. 122 p. 



5. 2 a. 78 p. 

6. 6 a. 2 p. 



7. 5 a. 42 p. 

8. 2 a. 151 p. 



Exercise VII. Page 274 

2. 98 a. 92 p. 



9. 
10. 

11. 



Exercise VIII. Page 277 

AE =3.75 ch. 4. AE = 5.50 ch. 

^LE = 3.50ch.; 5. CE = 4.456 ch. 

EG = SA2ch. 6> AD = 2.275 ch.; BE = 1.82 ch. 

AE = 4.55 ch. 7. AD = 4.51 ch.; BE = 3.61 ch. 

The distances on AB are 2 ch., 3 ch., and 5 ch. 

EM (on EA) = 2.5087 ch. ; AN (on AB) = 6.4390 ch. 

LetEG>DF; then AE = 12.247 ch., AG = 9.798 ch., AD = 8.660 

ch., ^=6.928 ch. 
LetDG>EF; then CG = 14.862 ch., CD = 13.113 ch., CF= 11.404 

ch., CE= 10.062 ch. 



Exercise IX. Page 295 

1. 9.5. 

2. Column H.G. 20.8, 20.4, 20.0, 19.6, 19.2, 18.8, 18.4, 18.0, 17.6, 

17.2, 16.8, 16.7. 
Column C. 0.0, 5.3, 6.4, 7.4, 5.0, 5.1, 6.2, 8.5, 6.0, 7.0, 7.2, 0.0. 




10 10.25 



ANSWERS 

5. Third column : 26.944 opposite ; 25.286 opposite 4. 

Fifth column : 20, 19.5, 21.3, 23, 22.3, 21.431, 20.4, 21.8, 24.1. 




Exercise X. Page 301 

1. 9986.5 cu. yd.; 9994.9 cu. yd. 

2. 5730ft.; 2865ft.; 1910ft.; 1433ft.; 1146ft. 

3. 1365 ft.; 4° 11' 53". 



NAVIGATION 







Exercise I. Page 


319 








1. 


S. 35° 56' E. 


7. N. 56° 22' E. 




13. 


S. 42° 


0'E. 


2. 


N. 77° 45' W. 


8. S. 73° 2(V W. 




14. 


N. 30° 


0' W 


3. 


S.E. by E. 


9. S. 39° 15' E. 




15. 


N. 52° 


0' W 


4. 


S. 79° 4' W. 


10. N.E. by E. 




16. 


S. 27° 


0' w 


5. 


N. 82° 47' W. 


11. N.N.W. 




17. 


S. 17° 


0' w 


6. 


S. 60° 30' E. 


12. W. 




18. 


N. 17° 


56' E. 



Exercise II. Page 332 

L" = 45° 26' N., p = 405.8 W. 

L" = 2°46'S., p = 405.6 E. 



3 L"= 0° 3'S., 


p = 406.0 E. 


4. L"= 2° 12' N., 


p = 52.1 E. 


5. L" == 36° 57' N., 


p = 113.1 W 



28 NAVIGATION 



6. 


C 


= 61° 49', 


P 


= 317.3. 


7. 


c 


= 25° 19', 


P 


= 151.4. 


8. 


D 


= 118.3, 


P 


= 23.1 W. 


9. 


B 


= 119.6, 


P 


= 57.9 W. 


10. 


L" 


= 37°28'N., 


D 


= 58.0. 


11. 


L" 


= 17°54'N., 


D 


= 72.2. 


12. 


C 


= 25° 27', 


P 


= 128.5. 


13. 


c 


= 22° 21', 


P 


= 308.8. 


14. 


L" 


= 6° 3'N., 


P 


= 95.4 E. 


15. 


L" 


= 14° 3' S., 


P 


= 137.5 E. 


16. 


L" 


= 40° 12' S., 


P 


= 239.7 E. 


17. 


C 


= S. 53° 22' W., 


D 


= 360.4. 


18. 


C 


= N. 56° 6'E., 


D 


= 154.2. 


19. 


c 


= N. 75° 5'E., 


D 


= 147.7. 


20. 


c 


=.N. 34° 20' W., 


D 


= 93.3. 



Exercise III. Page 335 

1. 505.4 E. 5. 80.1 W. 9. 172° 15' E. 

2. 125.0 E. 6. 166° 50' E. 10. 12° 47' W. 

3. 109.1 W. 7. 30°13'W. 11. 179° 11' E. 

4. 246.0 E. 8. 179° 48' E. 12. 21° 18' W. 

N. 58° 40' E. 







Exercise IV. 


Page 340 


1. 


C 


= E.N.E., 


D = 295.3. 


2. 


C 


= N. 61° 53' W., 


D = 212.2. 


3. 


c 


= S. 64° 34' W., 


D = 209.6. 


4. 


L" 


= 49°10'S., 


X" = 176° 0' W. 


5. 


L" 


= 18°52'N., 


X" = 175° 12' E. 


6. 


L" 


= 0°59'S., 


X"= 27°47'W. 


7. 


L" 


= 42°11'N., 


X"= 65°46'W. 


8. 


i" 


= 41° 56' N., 


X"= 63° O'W. 


9. 


X" 


= 41°36'N., 


X"= 59°12'W. 









ANSWERS 








10. 


X" = 57°55' W., 






1) =61.5. 






11. 


X" =61° 29' W., or 54 


°5' W. 


C = 87° 15'. 






12. 


C = S. 57° 19' E. 






D = 131.5. 






13. 


C =S. 71° 9'E. 






D = 510.7. 






14. 


C = S. 53 C 46'E., 






1) =289.3. 






15. 


C = N. 36° 25' W. 






D = 156.6. 






16. 


£" = 45° 10' K, 






X" = 69° 27' W. 






17. 


L" = 35° 14' N., 






X" = 73° 2' \V. 






18. 


L" = 42°24'N., 






X" = 66° 15' W. 






19. 


L" = 39°59'N., 






X" = 68° 35' W. 






20. 


Z" = 53°20'N., 






X" = 1° 25' W. 






21. 


L" = 30° 4'N., 






X" = 67° 27' W. 






22. 


L" = 33° 8'N., 






X" = 66° 19' W. 






23. 


L" = N. 64° 55' W. 


? 




D =668.5. 






24. 


£" = 25°42'N., 
D =807.9. 






C = S. 36° 27' W. 








Exercise V. 


Page 


347 


1. 


C 


= N. 


46° 29' E., 




D = 


: 140.9. 


2. 


c 


= S. 


30° 23' E., 




1) = 


: 98.5. 


3. 


c 


= S. 


28°24'E., 




D = 


: 261.5. 


4. 


c 


= N. 


55°13'E., 




D = 


: 2160. 


5. 


c 


= N. 


43° 39' W., 




D = 


: 248.8. 


6. 


L" 


= 47 


KTN., 




X" = 


: 32° 15' W. 


7. 


L" 


= 59 


° 3'N., 




X" = 


: 2° 43' E. 


8. 


L" 


= 47 


°20' N., 




X" = 


: 13° 45' W. 


9. 


C 


= N. 


33° 20' W., 




X" = 


:61 c 30' W., or 35° 10' W. 


10. 


X" 


= 23 


° 5' W., 




D = 


: 1022. 


11. 


C 


= S. 


51°59' W., 




D = 


: 358.8. 


12. 


C 


= N. 


16° 2'E., 




D = 


: 5599. 


13. 


C 


= S. 


60°29 , E., 




D = 


: 7685. 


14. 


\i 


= 4° 


48', 




D = 


: 476.3. 


15. 


By Mercator's, L" = 37° 


10' 


N., 


X" = 22° 47' W. ; 




Mid. Lat., L" = 37° 


16' 


N., 


X" = 22° 43' W. 



20 



30 NAVIGATION 



Exercise VI. Page 351 

1. L d = 0° 57' S., p = 21.4 W. 5. L d = 0° 4' S., p = 25.0 W. 

2. £ d = 1° 37' S., p = 45.6 W. 6. L d = 0° 15' lS.,p = 36.5 W. 

3. L d = l°39'S.,p = 44.6 W. 7. L d = 0° 7'S,p=6.8W. 

4. £ d =0°45'N.,;p = 37.1E. 8. L d = 0° 11' S., p= 110.7 E. 

9. Z d = l°2'S., p = 119.5 E. 





Exercise VII. 


Page 359 


1. 


J/' = 45 18'N., 


, \" = 19° 39' W. 


2. 


L" = 30° 49' N.. 


X'' = 17° 56' W. 


3. 


i // = 17°24 / S., 


X" = 1° 29' W. 


4. 


X // = 46°51 / N. 


, X" = 48° 6' W. 


5. 


X ,/ = 61°34 , N., 


, X" = 149° 49' E. 


6. 


L" = 51° 29' S., 


X" = 176° 35' E. 


7. 


X // = 36°31 , N V 


, X" = 2° 10' W. 



Exercise VIII. Page 374 

Course from the Bermudas = N. 49° 32' E. ; 
Course from the Lizard = N. 88° 42' W. ; 
Distance = 2812 miles ; L of V = 49° 59' N. ; 
X of V = 6° 54' W. 

Course from Boston = N. 55° 24' E. ; 
Course from Cape Clear = N. 77° 42' W. ; 
Distance = 2500 miles; L of V = 52° 28' N.; 
X of V = 25° 2' W. 

Course from Honolulu = N. 32° 3' E. ; 
Course from Vancouver = S. 50° 17' W. ; 
Distance = 2228 miles ; L of V = 60° 22' N. 
X of V = 80° 41' W. 

Course from Sandy Hook = N. 52° 50' E.; 
Course from Cape Clear = N. 78° 18' W. ; 
Distance = 2698 miles ; L of V = 52° 23' N. ; 
X of V = 24° 18 W. 



ANSWERS 31 

5. Course from Cape Frio = N. 23° 2' E.; 
Course from the Lizard = S. 34° 4' W. ; 
Distance = 4806 miles ; L of V = 68° 53' N. ; 

\of F = 57°26'E. 

6. Course from Cape Good Hope = S. 85° 13' W. ; 
Course from Cape Frio = S. 63° 23' E. ; 
Distance = 3208 miles ; L of V = 34° 37' S. ; 

X of V = 10° 3' E. 

7. Course from Perth = S. 87° 45' E. ; 
Course from Grand Port = S. 64° 37' W. ; 
Distance = 3158 miles ; L of V = 32° 8' S. ; 

X of V = 111° 29' E. 

8. Course from A = N. 43° 57' E. ; 
Course from B = S. 83° 51' W.; 
Distance = 3714 miles ; L of V = 48° 19' N. ; 

X of V = 3° 39' E. 

9. By Great Circle : 

Course from A = S. 49° 28' E. ; 
Course from B = S. 36° 32' W. ; 
Distance = 6772 miles ; L of V = 54° 24' S. ; 
Xof7= 158° 26' W. 
By Rhumb-line : 
Course from A = N. 76° 58' E. ; 
Course from B = S. 76° 58' W. ; 
Distance = 7432 miles. 







Exercise IX. Page 387 






1. 


25° 1'20". 


5. 56° 35' 28 ,/ . 


8. 


25° 57' 27' 


2. 


15° 10' 41". 


6. 60° 19' 5". 


9. 


20° 13' 41 


3. 


18° 27' 11". 


7. 31° 35' 13". 


10. 


35° 51' 13' 


4. 


30° 23' 4". 









Exercise X. Page 389 



1. 


July 


8 


7 6 10 p.m. 


2. 


Mar. 


8 


25 30 a.m. 


3. 


Jan. 


2 


6 10 10 A.M. 


4. 


Jan. 


1 


3 a.m. 


5. 


Feb. 


2 


8 4 30 p.m. 



6. June 30 23 8 25. 

7. Mar. 2 11 56 56. 

8. Aug. 31 10 8 20. 

9. Aug. 31 12 12 15. 
10. Dec. 31 22 41 56. 



32 



NAVIGATION 



Exercise XI. 

d. h. m. s. 

1. May 4 17 35 35. 

2. July 31 1 53 50. 

3. July 31 19 32 58. 

4. Mar. 2 6 57 0. 

5. Mar. 25 17 49 42. 



Page 391 





d. h. m. s. 


6. 


Dec. 31 6 7 a.m 


7. 


July 4 11 55 p.m. 


8. 


July 4 11 55 Op.ii 


9. 


May 20 7 43 20 a.m 


10. 


Jan. 1 3 16 40 a.m 



Exercise XII. Page 396 



1. 


22° 21' 28. 1" 


S.; 


2. 


17° 58' 4.7" 


N. 


3. 


16° 23' 7.9" 


N. 


4. 


16° 15' 35.0" 


N. 


5. 


22° 17' 40.7" 


S.; 


6. 


19° 55' 0.2" 


N. 


7. 


14° 29' 26.9" 


S.; 


8. 


7° 30' 22.7" 


S.; 


9. 


22° 46' 38.4" 


N. 


0. 


16° 23' 34.9" 


S.; 



6 32.79 to be added to app. time. 
6 6.95 to be added to app. time. 
3 28.48 to be taken from app. time. 

5 30.97 to be added to app. time. 

9 33.83 to be taken from app. time. 

6 14.30 to be added to app. time. 
16 18.84 to be taken from app. time. 

13 30.16 to be taken from app. time. 
1 25.08 to be added to mean time. 

14 4.88 to be taken from mean time. 



Exercise XIII. Page 404 



6° 48' 11" N. 

5° 26' 81" S. 

0° 3 , 44 // S. 

45° 43' 41" N. 

47° 13' 47" N. 
43° 40' 30" S. 



9. 
10. 
11. 
12. 



15° 58' 50' 
0° 9'53 y 
4° 31' 44 y 

49° 14' 14' 

5° 39' 37' 
6° 10' 58' 



N. 

N. 
N. 
N. 
S. 
N. 



13. 29° 24' 38" S. 

14. 0° 25' 35" S. 

15. 50° 19' 10" S. 

16. 7° 51' 0"N. 

17. 6° 31' 27" S. 



1. 53° 37' 33" N. 

2. 18° 15' 19" N. 

3. 34° 9' 18" S. 

4. 14° 36' 16" S. 

5. 60° 34' 36" S. 



Exercise XIV. Page 405 



6. 16° 14' 49" N. 

7. 50° 32' 48" N. 

8. 29° 34' 14" S. 

9. 52° 35' 28" N. 
10. 35° 41' 47" N. 



11. 45° 35' 1"S. 

12. 48° 28' 5"S. 

13. 46° 6'32"S. 



ANSWERS 33 

Exercise XV. Page 413 

1 . 16° 54' 0" E. 11. 44° 25' 45" W. 

2. 13° 53' 0" E. . 12. 76° 5' 30" E. 

3. 12°55 / 45" E. 13. 148° 17' 45" W. 

4. 10° 54' 30" E. 14. 179° 17' 45" W. at sights. 
5. 



49° 24' 45" W. 179 ° 46' 32" E. at noon. 



6. 41° 28' 15" W. 15. 34°45 / 0" W. 

7. 19° 68' 30" W. 16- 36 ° 51' 45 " W. 

8. 35° 26' 0" E. at sights. ". 45° 55' 0" W. 
35° 15' 11" E. at noon. 18. 47° 7' 0" W. 

9. 170° 58' 30" E. at sights. 19. 74° 21' 45" W. 
171° 17" 21" E. at noon. 20. 75° 16' 15" W. 

10. 93° 50' 45" W. 



LIBRARY OF CONGRESS 

llilll Hill III!! Illll IIIH Hill Mil Ml 



003 580 604 A » 



*— S:^; 






^^H 



:X'-'-- 



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